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I occasionally have the opportunity to argue with anti-Cantor cranks, people who for some reason or the other attack the validity of Cantor's diagonalization proof of the uncountability of the real numbers, arguably one of the most beautiful ideas in mathematics. They usually make the same sorts of arguments, so years ago I wrote up this FAQ to deal with them. Unfortunately, it's still hard to get anywhere with these people; the discussion frequently turns into something of this form:

ME: Suppose there is an ordered list containing all the real numbers. Then we can read off the diagonal entries and construct a real number that differs in the Nth decimal place from the Nth real number on the list. This real number obviously cannot be in the list. So the list doesn't contain all the real numbers.

THEM: Of course your proposed number is not on the list; it's not a well-defined real number.

ME: What do you mean? I gave you the exact procedure for constructing it. You take the Nth real number on the list, and you make it differ from that number in the Nth decimal place.

THEM: But if we really have a list of all the real numbers, then your proposed number has to be somewhere in the list, right?

ME: Yes, of course, so let's say it's in the 57th place. Then it would have to differ from itself in the 57th place, which is impossible!

THEM: Exactly, it's impossible! Your definition requires that it differs in some place from itself, which is impossible, so your definition is bad.

ME: But you're only saying that it's impossible on the basis of the assumption that there's a complete list of real numbers, and the whole point is to disprove that assumption.

THEM: But we're doing this proof under that assumption, so how can we make a definition that runs contrary to that assumption?

ME: But that definition is a good one regardless of whether there are countably or uncountably many reals. It is a complete, algorithmic, unambiguous specification of the real number. What else could you want?

THEM: I want the definition to be both unambiguous and non-contradictory, and your definition is contradictory!

ME: Forget about the purported complete lists of real numbers for a moment. Don't you agree that for any list of real numbers, complete or incomplete, it's possible to construct a real number that differs in the Nth place from the Nth number on the list?

THEM: No, it's only possible to construct such a real number if that real number isn't on the list, otherwise it's a contradictory definition.

ME: Don't you see that the contradiction is not the fault of my perfectly good definition, but rather the fault of your assumption that there are countably many real numbers?

THEM: No, I don't.

ME: But what if we took our putative complete list of real numbers, and fed it in line by line into a computer with an algorithm that spits out, digit by digit, a real number that differs in the Nth digit from the Nth number on the list? Would such a computer program work?

THEM: No it wouldn't, the computer program would hit the place on the list where the number being constructed would reside, and then it would get crash, because it can't choose a digit for the number that differs in the nth place from itself.

ME: Argh!

So how do I stop going in circles and convince them that they're wrong?

Any help would be greatly appreciated.

Thank You in Advance.

Keshav Srinivasan
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    If they are true cranks, give up. Ain't gonna happen. – Ross Millikan Oct 15 '13 at 15:30
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    @RossMillikan Well, there are a lot of people who aren't true cranks, and are genuinely just confused about Cantor's proof. They often have peculiar misconceptions (like an infinite set must have infinitely large members), but if you argue with them long enough you can often get through to them. Those are the kinds of people I was trying to help with my FAQ. – Keshav Srinivasan Oct 15 '13 at 15:36
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    I only discussed with anti-Cantorians after I had uttered a discouraging word to a student. Serving my penance, you see. – Jyrki Lahtonen Oct 15 '13 at 15:36
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    Your biggest problem with this hypothetical anti-Cantor crank is that E seems to not understand proof by contradiction. I guess there should be some overlap between Cantor-truthers and people who don't understand contradiction, but my experience has been that the problem is usually more along the lines of not understanding the relationship between "sets", "lists", and "countable". Discomfort with contradiction is a separate issue that should be straightened out in a calmer setting, not in a case where the result is so unintuitive. – Eric Stucky Oct 15 '13 at 15:41
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    Point out to them that, by their argument "we assume a list of all real numbers, so it's impossible to construct a number not on the list", you could also prove that the only color is black, since assuming that black is the only color we cannot find another one. If that doesn't work then they don't understand logic and will never be convinced of anything. – Ryan Reich Oct 15 '13 at 15:42
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    Let me just say that this is one of the best questions I've seen on M.SE. +1, but more in spirit! –  Oct 15 '13 at 16:05
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    You have to agree on a definition of "real number" before you can prove that they aren't countable. Otherwise your crank is right... how can you be sure that the string of digits you just made up is still a "real number", and that it's not equal to any other "real number" on the list? – mjqxxxx Oct 15 '13 at 16:32
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    @mjqxxxx Well, they do think that every infinite decimal specifies a real number, but they don't think Cantor's procedure validly specifies an infinite decimal, at least if the list is complete. They don't realize you can carry out this procedure for any list. – Keshav Srinivasan Oct 15 '13 at 16:41
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    Your imaginary person seems to have an issue with proof by contradiction in general, not with cantor's diagonalization argument specifically. If we assume X, Y and Z are true, and can prove that X is false, then one of our assumptions must be faulty. – BlueRaja - Danny Pflughoeft Oct 15 '13 at 19:29
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    @BlueRaja-DannyPflughoeft The problem is, they conclude that the assumption that's faulty is the validity of the construction of the "anti-diagonal", rather than the completeness of the list of real numbers. They don't understand that that construction must be valid for all possible lists, so the problem must be in assuming that a list of real numbers could be complete. – Keshav Srinivasan Oct 15 '13 at 20:40
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    `THEM: I want the definition to be both unambiguous and non-contradictory, and your definition is contradictory!` - They almost understood. Stop there and explain what a "proof by contradiction" is. – Izkata Oct 15 '13 at 21:14
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    I think that you may need to more clearly distinguish between the "Set of all Reals" and a "List of all Reals". The point of Cantor's proof is that there is such a set, but there cannot be such a list. – RBarryYoung Oct 15 '13 at 21:24
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    Have you read [What to do when the trisector comes](http://web.mst.edu/~lmhall/WhatToDoWhenTrisectorComes.pdf‎) by Underwood Dudley? – detly Oct 15 '13 at 22:33
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    The only way out is this: real numbers are not the same thing as natural numbers. The interval from, say, 0 to 1 is not a set which contains a list of things, but a continuum. Stop talking about lists. Real numbers are not countable because they are not computable (except for notable exceptions, such as ones which correspond exactly to some other kinds of numbers). For instance, we cannot distinguish whether some real number is pi, or whether is something very close to pi. A discrepancy can occur arbitrarily far in the decimal expansion, just beyond to where we quit looking. – Kaz Oct 16 '13 at 00:41
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    Of course, there are plenty of serious anti-Cantorians, Brouwer coming first to mind. At the heart, there is a very real disagreement to be had about whether an algorithm should be identified with a number. And the topos theorists know quite well that there are huge advantages to keeping the two ideas separate—in (classical!) relative geometry, for example, you can import constructive but not classical theory. But it's probably as pointless to argue intuitionism with the Cantor cranks as it is for them to try persuading us that it is possible to "carry out" a non-terminating procedure. – Slade Oct 16 '13 at 02:10
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    There is an error in step 3 of your proof: "$f(A)$ cannot be an element of the range of $A$, since it differs in the $n$th digit from the $A(n)$." The conclusion is correct but the reason is not, since decimal representations of real numbers are not unique. – user7530 Oct 16 '13 at 04:46
  • @Keshav: Would Paola Cattabriga be deemed an 'anti-cantor crank'? I am referring to her arXiv preprint "Beyond Uncountable" (arXiv:math/0312360[math.GM]). In this paper she claims that because the absolute complement of a set A (that is {x| x 'not a member of' A} cannot be defined [except as a proper class?] in ZF but rather, the complement of A can only be defined relative to another set. She further claims that because of this, Cantor's theorem |X| < |Powerset(X)| and his diagonalization argument do not hold in ZF. This strikes me as rather odd. However, looking at the proofs she gives, – Thomas Benjamin Oct 19 '13 at 17:32
  • they seem to make sense. Here is, (in a nutshell) her argument against diagonalization in ZF. Consider two lists, L, and its dual L*. L is a countable set of countably infinite binary sequences, L* formed from L in the following manner: given an arbitrary binary sequence s in L, s* in L* is formed by replacing every 1 in s by a 0, and every 0 in s by a 1. There seems no reason to assume L* is uncountable. Now form the binary sequence that diagonalizes out of L. By Cantor's construction that sequence (call it d) would seem to be, by definition of L*, a member of L*, and there is – Thomas Benjamin Oct 19 '13 at 17:50
  • no reason to assume that L* is anything but countable. What, if anything, is wrong with this argument? – Thomas Benjamin Oct 19 '13 at 17:53
  • @ThomasBenjamin First of all, why would d be an element of L*? Second of all, I agree with you that L* is countable, but how does that show that there's something wrong with Cantor's argument? – Keshav Srinivasan Oct 19 '13 at 18:18
  • Remember, to get d so that it 'disagrees' with each member of L (so that it is not a member of L) Cantor's argument requires that each 1 of L's diagonal must be switched to 0 and each 0 in the diagonal must be switched to 1 to get the countably infinite binary sequence that proves Cantor's theorem. Assuming the diagonal (call it D) of L is itself a member of L,by definition the dual d of D is a member of L*, which you have agreed is a countable set so in fact by constructing d to diagonalize out of L one does not get into an uncountably infinite set but in L* which you – Thomas Benjamin Oct 21 '13 at 19:52
  • agreed was countable. – Thomas Benjamin Oct 21 '13 at 19:54
  • @ThomasBenjamin What you're saying makes no sense. No one ever said that all the real numbers that are not in L are in L*, so showing that L* is countable doesn't help you at all. The point is, d is not in L. So no matter what list of real numbers you take, there's a real number not on the list. So you can't have a complete list of real numbers, so there are uncountably many real numbers. – Keshav Srinivasan Oct 21 '13 at 20:54
  • In fact, in order to get Cantor's diagonalization result, you might end up (if Cattabriga is correct) having to construct a d that is contained in neither L nor L*, and now you seem (possibly) to have d satisfy sets of conditions, analogous to forcing.... – Thomas Benjamin Oct 22 '13 at 10:37
  • @ThomasBenjamin As soon as you've shown that d is not in L, you're done, because you've shown that L isn't complete, and since L was arbitrary, it follows that you can't have a complete list of real numbers. Since the real numbers can't be put in a one-to-one correspondence with the natural numbers, there are uncountably many real numbers. What's the flaw in that? – Keshav Srinivasan Oct 22 '13 at 16:09
  • @Keshav: Consider the binary sequence d not in L, add d to L to form L'. Is L' uncountable? No, it is still countable. One can do this iterating process a countably infinite number of times and still the set L'^(omega) (and even the set L'^(omega+1) is still a countable set. The question is, how many times does one have to perform the iteration for the iterated set to be uncountable. I suppose one could say omega_1 times since omega_1 is supposed to be the first uncountable ordinal but that seems to presuppose the very thing you are trying to prove, doesn't it? – Thomas Benjamin Oct 23 '13 at 00:54
  • Also aren't all lists well-ordered sets? If uncountable cardinals and ordinals exist one should be able to have, in theory, uncountable lists. – Thomas Benjamin Oct 23 '13 at 00:57
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    @ThomasBenjamin The definition of a list is a well-ordered set of order type omega, i.e. a sequence indexed by the natural numbers. So the sets you're taking about, like L'^(omega+1), aren't lists. The point is, Cantor's proof shows that no well-ordered set of order-type omega can contain all the reals; any such set must not contain some d. But for any set X, a one to one correspondence between N and X is equivalent to a well-ordering of X of order type omega. Thus there's no one-to-one correspondence between N and R, which by definition means that R is uncountable. What is wrong with that? – Keshav Srinivasan Oct 23 '13 at 02:59
  • @Keshav: I guess my question to you is this: given that the operation of forming from any binary sequence s in L its dual s* in L* is a 1-1 correspondence between L and L*, can you construct a d that is contained in neither L nor L*? Also considering the iterate L'^(omega+1), what is its cardinality? I would think |L'^(omega+1)|= |omega|. By the way, why can't sequences be indexed by ordinals? – Thomas Benjamin Oct 23 '13 at 07:58
  • Also as regards the claim that there is no 1-1 correspondence between N and R, given that |R|=|Powerset(N)|, read Ms. Cattabriga's argument against that in ZF. I once believed as you did but after reading "Beyond Uncountable", I have my doubts, especially considering the existence of countable transitive models. Since the language of set theory is a countable language, doesn't it trouble you in the least that, assuming the validity of Cantor's arguments, one can do set theory without really using uncountable sets at all? – Thomas Benjamin Oct 23 '13 at 08:20
  • It's easy to construct a number that's neither in L nor in L*: just create a new list whose odd-numbered elements are the elements of L, and whose even-numbered elements are the elements of L*, and then diagonalize the new list. But I don't see the point of doing that, since L* is irrelevant to Cantor's proof. As far as sequences indexed by countable ordinals, there's nothing wrong with them, it's just that a list is by definition a sequence indexed by the natural numbers, and the reason that lists are important is that they're equivalent to one-to-one correspondences with N. – Keshav Srinivasan Oct 23 '13 at 12:47
  • As far as countable models of ZFC, that's just an artifact of the Lowenheim-Skolem theorem, which is a general property of first-order theories. But if we take second-order ZFC (which replaces the axiom schema of replacement with a single second-order axiom of replacement, or the weaker theory that replaces the axiom schema of separation with a single second-order axiom of separation), then all models of the set-theoretic universe are uncountable. – Keshav Srinivasan Oct 23 '13 at 13:03
  • let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/11182/discussion-between-thomas-benjamin-and-keshav-srinivasan) – Thomas Benjamin Oct 23 '13 at 18:43
  • @keshav:For some reason I am unable to log into chat. Until I resolve this issue, would there be any problem with us continuing this discussion here? – Thomas Benjamin Oct 28 '13 at 10:53
  • Yes, we can continue the discussion here. – Keshav Srinivasan Oct 28 '13 at 13:28
  • @keshav: Regarding your claim about constructing a binary sequence that is in neither L nor L*, consider the following (this holds for both finite and infinite binary sequences): consider the following binary sequences 01,10; 10,11--in the first case the diagonal is 00 (this set does not contain the diagonal), in the second case 11 (this set contains the diagonal). For the first case, add both the diagonal 00 and its dual 11. For the second case add the diagonal's dual 00. Now it is clear that the new sets now formed, not being (so to speak) 'square matrices', have no diagonal. – Thomas Benjamin Nov 01 '13 at 10:30
  • What to do? Well, one could add a special symbol p for 'placeholder' which has the property p=p* and add this symbol to the sets in order to form a 'square matrix'. When this is done, the sets now become (in the first case) 01pp, 10pp,00pp, 11pp; and in the second case the set becomes 10pp,11pp,00pp,01pp (to conform to your construction). Now for each set, try to diagonalize out of each set. Following Cantor, the diagonal that should not be in the first set is 01pp, but as you can see, that is already in the first set. In the second set, the diagonal that should diagonalize out of – Thomas Benjamin Nov 01 '13 at 10:52
  • that set is 00pp, but as readily seen, that sequence is already in the second set as well. As is easily seen (I think) this readily generalizes to all finite sequences. As for the infinite case (that is, a countable set of countably long binary sequences, there are still two subcases, one where the diagonal is contained the set, the other where the diagonal is not. To understand how the infinite case works, consider the closed interval [o,1] and the decimal expansion 0.000.... . How many 0's in the decimal expansion? Well, one might say 'countably many', but assume the decimal expansion – Thomas Benjamin Nov 01 '13 at 11:03
  • has omega many 0's. There is no reason one could not add an extra 0 to get omega+1 0's in the expansion (there will still be a countable sequence of 0's in the expansion-- so will adding a countable number of 0's to the expansion keep 0.000... a countable sequence of 0's). The same, therefore, must happen with a countably long binary sequence, also, the same when you add the placeholder p to get an infinite 'square matrix. So for the infinite case, first subcase, add both the diagonal and its dual and two placeholders at the end to form the 'square matrix' (one adds the diagonal because – Thomas Benjamin Nov 01 '13 at 11:21
  • in Cantor's diagonal argument, I believe one assumes the diagonal D is already in the set). As in the finite case, the diagonal D* that should diagonalize out of the set is still in the set. In the infinite case, second subcase, add the dual D* to the set and one placeholder--it is clear that when one forms the diagonal D* that should diagonalize out of the set, D* is still in the set. Similarly with your case, by adding the proper number of placeholders at the end to form the 'square matrix'. As for the original diagonal argument with a 'omega x omega' matrix, it is true that – Thomas Benjamin Nov 01 '13 at 11:36
  • that the element D* is not in the set, but all that (to my mind, at least) that shows is that D* is not in the set--nothing more. In other words, by adding D* to the set and the proper number of placeholders to form the 'square matrix' the new D* that is formed is still in the set. Can you show otherwise? Oh, by the way, it may be the case that the cardinality of set of reals is greater than the cardinality of the set of natural numbers--the diagonal argument and its variants do not actually show this – Thomas Benjamin Nov 01 '13 at 11:47
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    @ThomasBenjamin You're not making any sense. The decimal expansion of any real number has order-type omega, so if you added "place holders" at the end and made it of order type omega + 1, that would be nonsensical. – Keshav Srinivasan Nov 01 '13 at 12:00
  • @keshav: I'm not sure I understand why adding placeholders at the end of a decimal expansion of order type omega (and should we be talking of 'order types' here?) to make the decimal expansion of 'order type' omega+1 nonsensical. If you consider the decimal expansion a 'map' of the location of a point on the line segment represented by, say, the closed interval [0,1], by restricting the length of decimal expansions to omega you might run the risk of having 'gaps' in the interval [0,1] whereby points on the line segment in question have no numerical 'map' associated with them. Please show me – Thomas Benjamin Nov 02 '13 at 12:07
  • why my claim is nonsensical? If I am 'hearing' the subtext correctly, it sounds as if we are dealing with a definitional issue here. Are we? Suppose one wishes to add the diagonal not contained in the original list to the list in order to obtain a new binary sequence not in the list. That seems allowable, but by adding the diagonal to the list, wouldn't that list now be of 'order type' omega+1? Why is that not allowable? – Thomas Benjamin Nov 02 '13 at 12:11
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    @ThomasBenjamin If you put placeholders at the end of an infinite decimal expansion, then you'll no longer be working in the real number system, but rather in some some other system of numbers of your own invention, because all real numbers have decimal expansions of order type omega (or less). As far as putting a number at the end of your list, that's not allowed, because a list is by definition a sequence of order type omega. – Keshav Srinivasan Nov 02 '13 at 15:30
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    @ThomasBenjamin And the reason we care about sequences of order type omega is that every one-to-one correspondence between the natural numbers and a set X can be represented as a sequence of order type omega in X. So if we show that no sequence of order type omega contains all the real numbers, we'll have proven that the real numbers are uncountable. – Keshav Srinivasan Nov 02 '13 at 15:34
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    @ThomasBenjamin In fact, we can phrase the theorem slightly differently as follows: suppose that f is a function from N to R. Then there is a real number not in the range of f. When you phrase it like that, it's clear that you can't add a number to the "end". – Keshav Srinivasan Nov 02 '13 at 15:37
  • @keshav: given these strict parameters, of course the theorem follows. It is apparent, at least to me, that all this hinges on defining lists as strictly of 'order type' omega and the 'ordinal length' of a real number as being strictly of 'order type' omega. This, of course, is perfectly OK. Though this is (perhaps) in violation of a directive given by the System Lords (and Ladies?) in control of mathstackexchange, I wish to thank you for indulging me in this discussion. It has been very helpful and enlightening. – Thomas Benjamin Nov 03 '13 at 08:45
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    @ThomasBenjamin As I said, we're not just arbitrarily restricting ourselves to sequences of order type omega. The reason why sequences of order type omega are important is that it can be very easily proven that if X is countable, then there exists a sequence of order type omega which contains all the elements of X. (Would you like to see the proof? )So if we manage to show that no sequence of order type omega contains all the real numbers, then it will immediately follow that the real numbers are uncountable. – Keshav Srinivasan Nov 03 '13 at 10:03
  • @ThomasBenjamin Also, we are not just arbitrarily defining real numbers to have decimal expansions of order type omega. This is a fact that can be proved using any of the standard definitions of the real numbers; whether through axioms of a complete ordered field, through the Dedekind cut construction of the reals, or through the Cauchy sequence construction of the reals, it's provable for any real number x, there exists an integer m, such that x = m + the sum from n = 1 to infinity (i.e. a sum of order type omega) of a_n/10^n, where each a_n is between 0 and 9. Do you want a proof of this? – Keshav Srinivasan Nov 03 '13 at 10:10
  • @ThomasBenjamin In any case, even if you liked some larger system than the real number system, one with placeholders allowed at the end of infinite decimal expansions, if we show that the real numbers are uncountable, then it immediately follows that your larger system is also uncountable, because a countable set can't have an uncountable subset. (I can give you a proof of this too.) – Keshav Srinivasan Nov 03 '13 at 10:14
  • @keshav: please give proofs for all you said you can prove. Also, by way of clarification, are sequences of ordinal length omega+1, omega+2,..., omega^omega, epsilon, etc deemed to be of 'order type' omega, because each of these can be placed in 1-1 correspondence with sequences of order type omega (making omega just the cardinal number Aleph-null)? – Thomas Benjamin Nov 04 '13 at 06:39
  • @ThomasBenjamin Proof 1: Suppose X is countable, which by definition implies that there exists a one-to-one correspondence (AKA a bijection) f from N to X, so define a sequence (x_n) (of order type omega) by x_n = f(n). Then since f is a one-to-one correspondence, it follows that all elements of X are in the range of f, so (x_n) contains all the elements of X. – Keshav Srinivasan Nov 04 '13 at 07:34
  • @ThomasBenjamin Proof 2 is given here: books.google.com/books?id=25qHHYS310EC&pg=PA15&lpg=PA15&dq=real+analysis+decimal+expansion&source=bl&ots=HedRYgZsFl&sig=tsW6zOGcLJawYqW4Fp5GshhRQaw&hl=en&sa=X&ei=FU93UpKiMeLdsATpo4D4Aw&ved=0CEgQ6AEwBg#v=onepage&q=real%20analysis%20decimal%20expansion&f=false – Keshav Srinivasan Nov 04 '13 at 07:46
  • @ThomasBenjamin Proof 3 just involves an application of the Cantor-Schroeder-Bernstein theorem, proven here: en.wikipedia.org/wiki/Cantor–Bernstein–Schroeder_theorem#Proof As far as whether sequences of "ordinal length" other than omega have order type omega, the answer is obviously no; "ordinal length" is just another term for order type. It is true that sequences of any countable order type are countable, but to reiterate, the reason why we specifically care about sequences of order type omega, as opposed to sequences of other countable order types, is the result proved in Proof 1. – Keshav Srinivasan Nov 04 '13 at 07:52
  • @ThomasBenjamin Proof 1 allows us to immediately conclude that the real numbers are uncountable as long as we can show that no sequence of order type omega contains all the real numbers, so we don't even need to worry about sequences of other countable order types. – Keshav Srinivasan Nov 04 '13 at 07:55
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    I thought this site is for question and answers on math. Why not stay with this? Even if it feels like candy to get inspired to "beat the bullshit" / to feel better than "someone-stupid" (in this case "crank"). Vote to close! – Gottfried Helms Feb 19 '14 at 07:50
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    From the small amount of discussion I've had with "opponents to Cantorism", it seems that the main gripe they have is not with the proof that the cardinality of R is larger than the cardinality of N, but with the notion that cardinality is somehow an appropriate measure of "largeness" of sets. – Benjamin Lindqvist Dec 26 '15 at 19:33
  • Ie I don't think it's unreasonable to discard any "measure of largeness" that doesn't satisfy $m(A) > m(B)$ if $B$ is a strict subset of $A$. "But we lose so much great mathematics if we don't accept it!" doesn't really strike me as good enough reason. I'd love to understand both sides of this issue better though, so any clarifications would be appreciated. – Benjamin Lindqvist Dec 26 '15 at 19:38
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    @BenjaminLindqvist Well, the anti-Cantor cranks that I'm talking about here explicitly claim that there can exist a one-to-one correspondence between N and R. – Keshav Srinivasan Dec 26 '15 at 19:52
  • I posted as a separate question at http://math.stackexchange.com/questions/1589887/cardinality-as-size-of-a-set instead, feel free to comment there. – Benjamin Lindqvist Dec 26 '15 at 20:30
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    Suggested reading on this problem: [An editor recalls some hopeless papers](http://www.math.uni-hamburg.de/home/khomskii/ST2013/Hodges.pdf). – Massimo Ortolano Apr 30 '16 at 19:59

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The formulation "If $f:\mathbf{N} \to \mathbf{R}$ is an arbitrary mapping, then $f$ is not surjective" clearly fixes the original list of real numbers and sets aside the potentially combatitive issue of whether or not the list is all of $\mathbf{R}$.

Significantly, the argument is no longer by contradiction, but by direct implication: The diagonal procedure constructs a real number not in the image of $f$. Perhaps this may help circumvent the sense of double-talk presumably conveyed in first positing the existence of an enumeration of $\mathbf{R}$, then arguing that some infinite decimal is not on said list.

Andrew D. Hwang
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  • Yep, good answer. Furthermore, emphasize that the diagonalization procedure works in pretty much all the major foundational systems. It follows from the axioms. If they do not think the procedure if valid, fine, but in this case the onus is on *them* to build a new foundations capable of building bridges and landing planes safely, in which the diagonalization procedure doesn't go through. As has been pointed out here before, its all well and easy to point out perceived flaws in a foundations, but much harder to actually build one that avoids those perceived flaws. – goblin GONE Oct 15 '13 at 16:10
  • I've tried to phrase Cantor's proof so that it's no longer a proof by contradiction; see the part that starts with "Forget about the purported completed lists...". But they always respond with something along the lines of "But that list could potentially be a complete list of real numbers, so you have to keep the possibility in mind that the number being constructed may already be on the list, in which case the construction is ill-defined because a number cannot differ in the nth place from itself. So in order to show that it's well-defined, you have to first show that it's not on the list." – Keshav Srinivasan Oct 15 '13 at 16:17
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    @Keshav Srinivasan: I've never really understood why some people have such difficulty with proofs by contradiction, since so many arguments in everyday life have the same form. For example, a person accused of murder might argue that he couldn't have killed 'X' because he was at a dinner party with several witnesses when 'X' was killed, so the assumption of killing 'X' contradicts his being not being able to be in two places at the same time. The same if you're a child and your mother accuses you of taking some cookies, and you've been in school all day. – Dave L. Renfro Oct 15 '13 at 19:25
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    @DaveL.Renfro That last example is regrettably incorrect. Mothers do not reason by logic :-) – Thomas Oct 16 '13 at 00:46
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    @user8921 Bridges and planes can be made to work with 64 bit floating-point numbers. No "real" numbers are involved. The role of pi is played by an actor named `3.1415926535`, or similar. – Kaz Oct 16 '13 at 00:49
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    @Thomas This is because mothers are by definition always correct :) – Neal Oct 16 '13 at 01:22
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    @Kaz, very true, but on a finite stage, the role of Cantor's diagonal argument would be played by another actor with an uncanny resemblance to the infinitary Cantor diagonal argument. – zyx Oct 16 '13 at 04:31
  • @DaveL.Renfro Oh, but the examples you gave don't necessitate a proof by contradiction. They are examples of direct proof extended by the redundant "If we assume to the contrary we get nonsense." – superAnnoyingUser Apr 26 '14 at 21:38
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    @DaveL.Renfro That in't a proof by contradiction, it's a proof by contrapositive. "If I am the murderer, then I was not at the party", to which the contrapositive is "If I was at the party, I am not the murderer." But even in mathematics, contradiction is often confused with contrapositive. – Jack M May 01 '16 at 12:11
  • I love this formulation. Am I right in thinking the diagonalisation procedure is not arbitrary; it's a specific mapping? Is it possible to prove that there does not exist any surjective mapping $f:\mathbb{N}\to\mathbb{R}$. Not that I want to argument to fail (as everybody round here seems to think if you challenge it!); it's that I want it to be watertight. – samerivertwice Jun 22 '16 at 13:47
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    @RobertFrost: Here's a relatively explicit approach. Assume $f$ is a mapping from $\mathbf{N}$ to the set $X$ of real numbers in $[0, 1]$ whose decimal expansion contains only the digits $0$ and $1$. Write$$f(n) = \sum_{k=1}^{\infty} a_{n,k} 10^{-k},$$with $a_{n,k} = 0$ or $1$. Put $b_{k} = 1 - a_{k,k}$ ("the $k$th digit differs from the $k$th digit of $f(k)$"); the number$$x = \sum_{k=1}^{\infty} b_{k} 10^{-k}$$is in $X$, but not in $f(\mathbf{N})$. – Andrew D. Hwang Jun 22 '16 at 22:12
  • @AndrewD.Hwang you've proven the part I already accept and like! But this is a specific mapping. I was asking if it can be proven that there does not exist a surjective mapping; i.e. this mapping fails but can it be proven that all mappings fail. – samerivertwice Jun 23 '16 at 07:11
  • @RobertFrost: Not sure I understand what you're asking, then. The preceding argument shows (constructively, for some value of "constructively") that the image of an _arbitrary_ mapping from $\mathbf{N}$ does not even contain $X \subset \mathbf{R}$, much less contain $\mathbf{R}$ itself. That is, if $f:\mathbf{N} \to X$ is _arbitrary_, the procedure constructs (algorithmically, based on $f$) the digits of a number $x$ in $X$ but not in $f(\mathbf{N})$, so that $f$ is not surjective. – Andrew D. Hwang Jun 23 '16 at 08:52
  • @AndrewD.Hwang I can likewise create a mapping $f:\mathbb{N}\to\mathbb{N}$ as follows $f=6x$ which as you say does not even contain $G\subset \mathbb{N}$ where $G$ is say the set of even numbers. But just because this *specific* mapping fails to enumerate $\mathbb{N}$, it does not mean an enumeration is not possible. By your argument I could argue $\mathbb{N}$ does not enumerate $\mathbb{N}$. – samerivertwice Jun 23 '16 at 13:48
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    My argument is fundamentally different: If you specify an _arbitrary_ mapping $f:\mathbf{N} \to X$, I'll give you an element of $X$ not in the image of your $f$. It follows that the arbitrary mapping you specified is not a surjection. That is, I still don't see your point. Could you perhaps clarify in what sense the "$f$" in my argument seems "specific"...? (The proof certainly _uses_ $f$, but that's to be expected; in order to construct a number not in the image of $f$, one has to refer to $f$ somehow. That fact doesn't preclude $f$ being arbitrary.) – Andrew D. Hwang Jun 23 '16 at 21:19
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An error of tactics and substance, made in that FAQ and an uncountable number of online debates of these matters, is to equate

reasonable objections to parts of the framework (including objections identical to ideas published and developed by accomplished mathematicians)

with

mistakes in digesting the proof on its own terms.

The first category, of coherent self-consistent criticisms that in some views or formalizations are correct objections, include

  • there can be no actual or completed infinity
  • proof by contradiction and/or excluded middle logic, is bad
  • there should be an effective procedure/definition for every number
  • the number of effective procedures/definitions is countable

You cannot overcome these criticisms as such. Instead, the explanation is to present Cantor's proof in a way that is compatible with the criticism either by showing that the disputed concept does not appear in the proof, or formulating the diagonalization argument as it would be stated in a finitist, constructive, predicative, computable, or definable mathematics.

zyx
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  • Well, in the FAQ I do discuss one legitimate philosophical objection, but I just say something like "If you don't accept ordinary Platonistic mathematics in the first place, then Cantor's theorem is about fictitious things anyway, so there's no need to dwell on it." The problem is that most people who take issue with Cantor's proof are not motivated by some coherent philosophical principle, they just have some basic misconception about mathematics, like "infinitely large set of numbers must contain infinitely large numbers". But I respect, e.g., legitimate predicativists: tinyurl.com/nelpa – Keshav Srinivasan Oct 16 '13 at 01:16
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    The FAQ item 1 is incorrect as it implies that definable reals (or sequences) are countable. This is only true in a non-definable mathematics. If you take the FAQ's suggestion to use definable objects only, then Cantor's theorem is still correct, but one has to formulate it for a definable world. It also is not true that not accepting Platonism or infinite sets invalidates diagonalization. – zyx Oct 16 '13 at 01:27
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    The thing is, the people the FAQ is intended for are not spending their energy arguing against some narrow version of Cantor's proof about effective enumerations of computable real numbers, for instance. They're arguing against the standard proof for the Platonistic result, and they're not realizing that they reject the presuppositions which even the statement of the theorem are based on. They're willing to accept that the statement of the theorem is meaningful, they just think it happens to be false, so I'm trying to say that that's not a coherent viewpoint. – Keshav Srinivasan Oct 16 '13 at 01:36
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    It is not likely you can formulate an unassailable argument that will succeed in every case. However, where an objection consists of coherent and incoherent parts, removing the dispute about the coherent part reduces the number of things in contention, and might help clarify to the skeptic that the incoherent part is not correct. Besides that, it is misleading both to the objector and to any observers should it be done in an online forum, to not handle the coherent part as being valid. – zyx Oct 16 '13 at 01:42
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    Thank you for standing up for reason. I am baffled by the number of people in mathematics who will loudly defend straw-man tactics when they are aesthetically favorable. It would make as much sense to ban anyone without a Ph.D. from discussing the diagonalization proof, on the grounds that they do not truly understand the foundations. As beautiful as Cantor's work is, I think that what has been done with topoi is even more so. It is thus a good thing, as you point out, that we do not really have to choose—at least so long as we don't get carried away by unfair and oppositional attitudes. – Slade Oct 16 '13 at 02:26
  • While the points you list are valid points of view, they should not be objections to the diagonalisation argument that is used to prove "For any set $X$ there is no bijection $X\to\mathcal P(X)$" or its positive counterpart (which forms its actual proof) "For any map $f:X\to\mathcal P(X)$ there exists $Y\in\mathcal P(X)$ that is not in the range of $f$". Note there is nowhere (in statement or proof) talk of infinity, nor of effective procedures, and the argument does not use excluded middle (it is intuitionistically acceptable). The proof of the negative form inevitably does use contradiction. – Marc van Leeuwen Feb 19 '14 at 08:11
  • @zyx I love this answer, it really hits the nail on the head but I would qualify "proof by contradiction and/or excluded middle logic, is bad." It's not bad *per se*, it's entirely rigorous within a logical system which is proven consistent. It's only within a system not proven consistent that proof by contradiction is not a proof. – samerivertwice Jun 22 '16 at 14:29
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    @RobertFrost That is nonsense. If the system is inconsistent then everything is true in that system and hence proof by contradiction is certainly valid. – Tobias Kildetoft Jun 27 '16 at 06:53
  • Consistency does not affect the validity of proof as a derivation but it does affect whether valid derivations become a source of certainty or of new knowledge. @TobiasKildetoft – zyx Jun 27 '16 at 20:31
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    @TobiasKildetoft We're into semantics I think; which is no bad thing! I wrote *a system not proven consistent*, not *a system known to be inconsistent*. What I mean by this is that in a system not proven consistent, the proof that something is false does not exclude the possibility that it is true. Therefore no truth and falsehood of the same fact can be deduced to be a contradiction arising out of {the supposition being tested} being incorrect. Am I mistaken in this notion? – samerivertwice Jun 30 '16 at 09:28
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You could try limiting the discussion to the finite case of Cantor's theorem as a first step. Show them that for every function $f$ on the finite set $\{1,\ldots,n\}$ there is a subset of the domain $\{1,\ldots,n\}$ that is not an element of $\text{ran}(f)$. Show them how the construction works for some examples, say $n = 2$ and $n=3$.

If they don't accept this argument in the finite case, then challenge them to write down a counterexample $f$. If they do accept it, ask them to point out what goes wrong when $\text{dom}(f)$ is $\mathbb{N}$ (or an arbitrary set, although this may be too abstract for them.) At least you might be able to separate their confusion about diagonalization from their confusion about infinity.

EDIT: I am talking about the version of Cantor's theorem for sets of natural numbers rather than the version for real numbers. If they do not understand the correspondence between real numbers and sets of natural numbers, their notion of "real number" is probably not precise enough to have a reasonable discussion about Cantor's theorem.

Trevor Wilson
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    These people usually don't know much if any set theory, so you can't really go through the proof of Cantor's theorem that the power set has a bigger cardinality with them. And in any case they're usually constructivists of some stripe, so they probably wouldn't accept the existence of the power set of N. – Keshav Srinivasan Oct 15 '13 at 16:22
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    @KeshavSrinivasan I appreciate your point, but I was hoping that they would be able to deal with the set theory in the finite case. Also, I purposefully phrased my answer so as not to mention the power set of $\mathbb{N}$, and not to rely on the power set axiom at all. If they don't accept the collection of all subsets of $\mathbb{N}$ as a _class_, then I don't see what they could possibly construe the real numbers to be. – Trevor Wilson Oct 15 '13 at 16:27
  • They think of real numbers, and natural numbers, as primitive objects, not defined in terms of set theory. You know, points on a number line. – Keshav Srinivasan Oct 15 '13 at 16:30
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    @KeshavSrinivasan Then maybe they are right, and there really are only countably many points on the number line ;-) You have to commit to some formal definition of the real numbers for Cantor's theorem to be a theorem, and if they understand this formal definition, then they should be able to understand how real numbers correspond to sets of natural numbers. If they don't, then you have a more fundamental problem to deal with first. – Trevor Wilson Oct 15 '13 at 16:34
  • Well, they're willing to accept that real numbers can be expressed as infinite decimals, but not much more. And they don't generally believe in the existence of arbitrary abstract sets. – Keshav Srinivasan Oct 15 '13 at 16:38
  • @KeshavSrinivasan If they don't believe in the notion of arbitrary sets of natural numbers, then their understanding of infinite decimal expansions must be flawed. What if they believe that all infinite decimal expansions have some "pattern"? – Trevor Wilson Oct 15 '13 at 16:55
  • It doesn't matter if you only believe in computable real numbers. Cantor's proof shows that you can't have an effective enumeration of the computable real numbers. But in any case, I think they do believe in arbitrary infinite decimals, just not in abstract sets. – Keshav Srinivasan Oct 15 '13 at 17:06
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    @KeshavSrinivasan But what if they are requiring the real numbers to be computable, but are not requiring the function $f : \mathbb{N} \to \mathbb{R}$ to be computable? My point is just that if you expect to convince them that you are right and they are wrong, then first you should fix precisely a context where this is true, and make sure they know what it is. If they have a precise understanding of "infinite decimal", they must also have a precise understanding of "set of natural numbers". – Trevor Wilson Oct 15 '13 at 17:11
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    ...I'm not saying they need to accept arbitrary sets; we can restrict our attention to the case where the domain of $f$ is either a natural number or $\mathbb{N}$ itself. – Trevor Wilson Oct 15 '13 at 17:11
  • Brouwer: '[the applicability of LEM] was caused historically by the fact that, first, classical logic was abstracted from the mathematics of the subsets of a definate finite set, that, secondly, an a priori existence independent of mathematics was ascribed to the logic, and that, finally, on the basis of this supposed apriority it was unjustifiably applied to the mathematics of infinite sets.' The Continuous and the Infinitesimal, J L Bell –  Jul 19 '14 at 12:35
  • I don't think the people who fail to accept it for the finite case aren't your usual "cranks". They're more likely people who don't understand the argument. Most "anti-Cantor cranks", being in the company of Kronecker, Poincare, Wittgenstein; just means they maintain a very onerous threshold before accepting something as truth. – samerivertwice Jun 22 '16 at 14:27
6

The conversation seems almost equivalent to this hypothetical one about the irrationality of $\sqrt 2$:

ME: Suppose there are positive integers $m$, $n$ such that $m^2=2n^2$. Then we can generate a list of all prime factors and read off the exponent of $2$. This exponent should be even (because it is in the unique factorization of $m^2$) but also odd (because it is in the unique factorization of $2n^2$) so the equalilty cannot hold for any $m,n$.

THEM: Of course your proposed exponent cannot exist; it's not a well-defined integer.

ME: What do you mean? I gave you the exact procedure for constructing it: factorize that number. Check the factorization theorem.

THEM: But if we really have a list of prime factors then factor 2 has to have some exponent, right?

ME: Yes, of course, so let's say it's 57. Then it is odd and it would be also the double of an integer, that is impossible.

THEM: Exactly, it's impossible! Your definition of the exponent requires that it is both odd and even, which is impossible, so your definition is bad.

ME: But you're only saying that it's impossible on the basis of the assumption that the equality $m^2=2n^2$ holds for some positive integers $m$ and $n$, and the whole point is to disprove that assumption.

THEM: But we're doing this proof under that assumption, so how can we make a definition that runs contrary to that assumption?

ME: But that definition is a good one regardless of whether the assumption is true or false. It is a complete, algorithmic, unambiguous specification of an integer (the exponent of 2 in the unique prime factorization). What else could you want?

THEM: I want the definition to be both unambiguous and non-contradictory, and your definition is contradictory!

ME: Forget about the equality to be disproved for a moment. Don't you agree that for any number it's possible to construct a prime factorization and define the exponent of $2$?

THEM: No, it's only possible to define such an exponent if the number $a$ to be factorized is not such that $a=m^2=2n^2$, otherwise it's a contradictory definition.

ME: Don't you see that the contradiction is not the fault of my perfectly good definition, but rather the fault of your assumption that $m^2=2n^2$?

THEM: No, I don't.

ME: But what if we took our putative number $a=m^2=2n^2$, and fed it into a computer with an algorithm that spits the exponent of $2$ in the prime factorization? Would such a computer program work?

THEM: No it wouldn't, the computer program would hit the place on the list where the exponent of $2$ would be computed and then it would get crash, because it can't produce a number that is both odd and even.

Marco Disce
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  • Interesting, can you spell out the proof that the square root of 2 is irrational using the exponent of 2? I've never seen that proof. I've only seen the conventional proof where you assume it's in simplest form and get a simpler form, or equivalently you construct an infinite descending sequence of natural numbers by continually dividing by 2. – Keshav Srinivasan Apr 30 '16 at 20:08
  • But yeah, you seem to have constructed an analogous situation to the one I keep encountering with anti-Cantor cranks. So how would you explain to them the error in their reasoning in your case, and in my case? – Keshav Srinivasan Apr 30 '16 at 20:09
  • It's very simple: $m^2$ must have an even exponent of $2$ in the prime factorization, $2n^2$ must have an odd exponent of $2$ – Marco Disce Apr 30 '16 at 20:14
  • OK yeah, that makes sense, thanks. So how would you respond to the "them" in the dialogue? – Keshav Srinivasan Apr 30 '16 at 20:18
  • I don't know if anti-Cantors are also anti-irrational numbers, the purpose of this parallel dialogue was to make people understand that there is a mistake in the anti-Cantor arguments assuming they think the proof of irrationality of $\sqrt{2}$ is correct – Marco Disce Apr 30 '16 at 20:19
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    Here also you can argue contrapositively, without making the initial controversial hypothesis: Let $m$ and $n$ be arbitrary positive integers. Each of $m^{2}$ and $2n^{2}$ factors uniquely into primes. However, the exponent of $2$ in $m^{2}$ is even, while the exponent of $2$ in $2n^{2}$ is odd. Since no integer is both even and odd, the factorizations of $m^{2}$ and $2n^{2}$ are distinct, and consequently $m^{2} \neq 2n^{2}$. Needless to say, I prefer this to "descent-type" arguments. :) – Andrew D. Hwang May 01 '16 at 13:39
4

It's not a proof, but in this situation it is reasonable to be like cranky student. When I say that his algorithm is wrong he can't agree until I show him any counterexample. So you can just ask your opponents to give you method of constructing such list. And you would always be able to show that list is incomplete:)

Anyway there are several different ways to prove that set of real numbers is uncountable. Are all of them rejected by your cranks?

Smylic
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    Ok, then you are just lucky they don't claim that set of integers is finite:) – Smylic Oct 16 '13 at 01:03
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    Well, sometimes the cranks have a claim that they have constructed a complete list of real numbers, but a lot of the time they just reject Cantor's claim that such a list cannot exist, as opposed to making an affirmative claim on their part. Or they cite a reason that the real numbers are countable, but that reason does not actually give you a procedure for making a complete list of real numbers. And yes, there are other proofs of uncountability, but for instance you need a fair amount of math background for the proof that a perfect set is uncountable, whereas Cantor requires little. – Keshav Srinivasan Oct 16 '13 at 01:03
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    I was 13 when I got to know about 5 different proofs and had understood all of them. I think the following is not too hard. Consider a numbered list of all real numbers of $[0, 1]$. Let cover $i$th of them by open interval of length $3^{-i}$. Since each point is inside its own inteval (which has positive length) then is should be covered by at least one inteval. But cumulative length of all invervals is $\frac12$, so union of them all can't cover all points of $[0, 1]$. – Smylic Oct 16 '13 at 01:13
  • Why wouldn't that proof show equally well that there are uncountably many rational numbers between $0$ and $1$? – Keshav Srinivasan Oct 16 '13 at 05:12
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    Not every point of $[0, 1]$ is rational, so covering every rational point doesn't mean covering every point and the whole segment. – Smylic Oct 16 '13 at 15:32
  • You're right, my point about uncountably many rational numbers doesn't make sense. But let me ask you this: if we take $[0,1]$ as a subset of R, and then cover every rational number with an open interval of length $3^{-i}$ (for the $i^{th}$ rational number), then can you give me an example of an irrational number that is not covered by any of those intervals? The rational numbers are dense in the reals, so it's hard for me to envision such an irrational number. – Keshav Srinivasan Oct 16 '13 at 18:43
  • Thank you for interesting questions in this discussion! Give me your numbering and I'll try to find such real:) Surely it exists. Considering fact that for any real $x$ and for any $\epsilon > 0$ there exist rational $q$ such that $|x - q| < \epsilon$. But for any of numbering of ratinals there should exist such $x$ that for any rational $q$ we have $|x - q| > \frac{3^{-n(q)}}2$. The closer to $x$ rational $q$ is, the shorter the interval it is covered by. No contradition. – Smylic Oct 16 '13 at 18:56
  • More precisely this algo gives you such real $x$. I claim that such $x$ exists in the iterval $[a_i, a_i+3^{-i}]$ of length $3^{-i}$, where $a_0=0$ and $a_i$ for $i > 0$ is defined as following. Open interval covering $i^{\text{th}}$ rational number divides interval $[a_{i-1}, a_{i-1}+3^{-i+1}]$ into at most two closed parts of cumulative length at least $3^{-i+1} - 3^{-i} =2\cdot 3^{-i}$. So at least one of these parts has length at least $3^{-i}$. Let $a_i$ be the least point of this part. Then $x = \lim_{i\to+\infty} a_i$ is never covered by any interval. BTW it works for any countable set. – Smylic Oct 16 '13 at 19:24
  • @KeshavSrinivasan you are misunderstanding the issue most people have with the diagonal argument, which is that it is a proof by contradiction and therefore there can be no 'misattribution' of the source of the contradiction. It simply is not a proof unless it includes a proof that the contradiction does not arise elsewhere; such as from the fact that at least one element of the 'diagonal' set is defined by the maximal element of some infinite set; a set which has no maximal element. Therefore the notion of the diagonal set is badly defined independently of the comparison of the cardinalities. – samerivertwice Jun 23 '16 at 20:57
  • @RobertFrost If you would like to try to get a contradiction without the assumption that the real numbers are countable, be my guest. In any case, what do you mean by "at least one element of the 'diagonal' set is defined by the maximal element of some infinite set"? – Keshav Srinivasan Jun 23 '16 at 21:00
  • @KeshavSrinivasan I don't have to get a contradiction without that assumption for cantor's diagonal argument not to be a proof. A proof by contradiction requires proof that the theory in which it's built is consistent in order to be a proof. – samerivertwice Jun 24 '16 at 08:22
  • @KeshavSrinivasan what I mean by "at least one element of the 'diagonal' set is defined by the maximal element of some infinite set" is that building the set $s$ by taking the complement of each element in turn can only build $s$ in its entirety if it does enumerate the enumerations, but it doesn't, it's a *specific* example of an enumeration that fails. So one can't proceed to the deduction that all means of enumeration will fail because the argument relies upon this specific means of enumeration, which happens to fail to create $s$. – samerivertwice Jun 24 '16 at 08:29
  • @KeshavSrinivasan If the argument included proof that the specific enumeration of infinite sets used in the argument could be generalised to all enumerations and therefore that all enumerations fail; that would be a proof. – samerivertwice Jun 24 '16 at 08:30
3

Instead of entering a discussion about the proof for real numbers, I would propose discussing instead the abstract version, which gives far less opportunity for polemic. Let the "crank" decide what he thinks about the abstract version and its proof, and then see where to go from there. If he refuses the abstract version, he should either be able to find fault with the proof, or accept being inconsistent (if finding no fault in the proof but still rejecting the conclusion). When accepting the abstract version one can still refuse application for the real numbers (for instance by denying the existence of infinite sets, or maybe accepting existence of the natural numbers but not of their power set), but it forces one to draw a line somewhere; once this is done there is really nothing left to discuss. For reference, here it is:

Proposition. For every set $X$ and every map $\def\P{\mathcal P}f:X\to\P(X)$ there exists $Y_f\in\P(X)$ such that for all $x\in X$ one has $Y_f\neq f(x)$.

Proof. Take $Y_f=\{\, z\in X\mid z\notin f(z)\,\}$. For $x\in Y_f$ one has $x\notin f(x)$ so $Y_f\neq f(x)$. For $x\in X$ with $x\notin Y_f$ one has $x\in f(x)$, so $Y_f\neq f(x)$. This establishes $Y_f\neq f(x)$ for all $x\in X$.

Marc van Leeuwen
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    The problem is, these people usually don't know much if any set theory, so you can't really go through the proof of Cantor's theorem that the power set has a bigger cardinality with them. Cantor's diagonal proof of the uncountability of the reals is usually the only proof they can comprehend. – Keshav Srinivasan Feb 19 '14 at 15:13
  • @KeshavSrinivasan: Apperently you are complaining it is just another proof they _cannot_ comprehend. Note that the only "set theory" really used is that one can define a subset by telling which elements to keep and which ones not to keep. You can view sets as lists of their elements if you like, and get close to the traditional diagonal, without having to bother about reals. I think if they cannot comprehend maps from a set to its power set, there is really no point in discussing countability of real numbers with them in the first place. – Marc van Leeuwen Feb 19 '14 at 15:30
  • If they learned enough set theory, they'd probably abandon their objections to Cantor's diagonal proof anyway. But in any case, I suppose the objection I described in my question could be carried over to the Cantor's theorem case: they would say that the definition of $Y_f$ doesn't make sense in the case when $f$ is surjective, and that the contradiction you get in the case when $f$ is surjective indicates that $Y_f$ does not exist, not that $f$ is not surjective. The fundamental problem with these people is that they're misattributing the source of the contradiction. – Keshav Srinivasan Feb 19 '14 at 15:39
  • Usually they don't know enough set theory to even understand power sets. But I find that it's still productive talking with (at least some of) them, because if you can resolve their confusion.concerning Cantor's diagonal proof, then very often it leads them to want to learn more about set theory and transfinite numbers. That's reward enough for arguing with them. – Keshav Srinivasan Feb 19 '14 at 15:47
  • Unfortunately, this proof doesn't hold for all set theories. It relies on restricted comprehension, the same axiom that doesn't allow you to have an universal set. I don't think anything would really stop you from having a set theory where you can find natural and real numbers and an universal set; in that theory the first diagonal argument could still hold, but the one regarding power sets wouldn't. (=drawing the line at the set proof is actually drawing the line higher than you would need to to accept the different cardinalities of natural and real numbers) – byserpas May 07 '15 at 18:01
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    @byserpas I don't understand at all what you want to say. It is not that restricted comprehension forbids you to have a universal set; rather it would require _another_ axiom, namely unrestricted comprehension, to have a universal set. But then you get the Russell paradox and an inconsistent theory. Also any diagonal argument is essentially about a power set or something very similar like digit strings (mapping natural numbers to $\{0,1,\ldots,9\}$ is very similar to mapping them to $\{\mathrm{false},\mathrm{true}\}$). – Marc van Leeuwen May 08 '15 at 06:03
  • Well, you can always have even more restricted comprehension (or differently restricted). Going to extremes, think of a theory where the only set that exists, All, contains itself and nothing more. Its only power set would be {{},{All}}... if {} were a set. but in this theory there's no empty set, just a set containing itself. so P(All) would be {{All}},if {{All}} were a set of this theory... is it? Well, its only element is All, so if we adopt extensionality, sure. P(All)=All. Your Y, in here, doesn't exist (as basically anything other than All), since it would be {}. – byserpas May 08 '15 at 06:44
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    @byserpas If there is only one set, then you've got no set theory (in the sense that you can only formulate statements that are trivially true or false). You need no other axioms than something "there is no $x$ but $A$, and $x\in A$ if and only $x=A$". You won't have any constructions like powerset (though you can give any name you like to the construction that always produces $A$), no natural numbers, no reals, no diagonal argument, nothing. It is an utterly boring enterprise. Go try to sell it to somebody as set theory, but don't waste our time with it. – Marc van Leeuwen May 08 '15 at 08:47
  • "All" is indeed a boring place, but I meant to offer it just as an example. It' not easy to describe a better example, so I'll provide a parallel. x^2=-1 doesn't have solution in R, but it has in C. #P(A)=#A doesn't have solution in ZFC, but it can have solution in other theories (I tried to provide an example, although boring). Suppose we have such a theory, and suppose it has enough axioms to guarantee the existence of reals and naturals. Are we supposing something impossible? Personally, I can't say for sure. If I wasted anyone's time, sorry, I won't anymore unless explicitly prompted. – byserpas May 08 '15 at 10:05
2

I think the only way would be to find a different proof that holds against the main concern of the typical objections:
Does the constructed number actually exist?

I'll try to illustrate a proof that can address this concern.
Let's consider lists of real numbers written in base 10 and build a number from the list using this rule:

If the n^th digit of the n^th number of the list is 1, we change it to 2, otherwise we change it to 1.

We're assuming the list is made of infinite rows; if not, we could always repeat the same numbers from the beginning in the same order and get our new number.
Let's give a name to numbers that can be obtained from lists with this particular rule: if I wanted to be specific, I would call them something like "(1->2)(else->1) diagonals", but for the sake of brevity i'll call them diagonals.
Any real with some digit different than 1 or 2 is not diagonal.
If a real only has ones and twos, it could be diagonal.

Note that if reals are countable, both diagonals and not diagonals are countable (as you could just list them in the order you found for the reals)

Now, what if we listed all the not diagonals? From where we are, we couldn't immediately conclude that the not diagonals are uncountable, since the diagonal number of the list of course wouldn't be in the list of not diagonal numbers.
But it exists: it's something made of just ones and twos. probably more ones.
So, there are diagonal numbers: you can easily imagine them in this setting.

But, what if we listed all diagonal numbers?

It's a list, so it has a diagonal number, not in the list by construction.
But it's a list of diagonals: it should have its diagonal number. So, usual argument, usual contradiction, just in a slightly different setting: in here, you can conclude that diagonal numbers aren't countable, but you can't say the same about not diagonal numbers. (at least, not immediately) But it doesn't matter, because the real numbers are made of diagonals and not diagonals, and if diagonals are not countable, neither are real numbers!

So, my hope is that this argument could break the chain a bit, since it does provide a list of real numbers that doesn't contain its diagonal (the not diagonals), which you don't need to prove are uncountable, but also a list of real numbers that, if it existed, it couldn't contain its diagonal nor not contain it.
Basically, good cop and bad cop i guess? Except with lists.

byserpas
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  • I think they would still make the same objection I discuss in my question: they would say that it's possible to construct a diagonal for the list of all non-diagonals, but it's impossible to construct a diagonal for the list of all diagonals. And their reasoning would be that if the diagonal for the list of diagonals did exist, then it would have to differ from itself, which is impossible. Therefore, the list of diagonals can't have a diagonal. The fundamental problem is that they don't realize that taking the diagonal is a valid operation for all lists. – Keshav Srinivasan May 07 '15 at 20:23
  • Well, the arguments in the essence are the same, so yes, it's likely things wouldn't change at all. But, sometimes a minimal superficial change can trigger a different reaction, even if the math behind it is the same... I mean, the reasoning to get is that one, there's no way around it. One can only arrange the same arguments in a different fashion, so that the counterarguments can fall in a different way until they hopefully unclick by themselves. – byserpas May 07 '15 at 20:43
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Cantor's proof is not about numbers at all (there is a problem with trying to use it on numbers), it is about infinite strings of two characters. See http://www.logicmuseum.com/cantor/diagarg.htm.

Call them Cantor Strings. Cantor used "M" and "W" which, while they are visually interesting, are hard to distinguish. You can use "0" and "1" and claim they represent fractions in binary representation, but there is a problem. For example, 5/16 has two such binary representations: 0.01010000... and 0.01001111... . So you can't claim a number is missing, if all you prove is that a string is missing.

Ignoring this fault, as most treatments do, your very first statement is not what Cantor does "ME: Suppose there is an ordered list containing all the (Cantor Strings)." The first step of Cantor's Proof is "Consider ANY list that links every natural number to a unique Cantor String." Nowhere is it assumed that this means all Cantor Strings.

In fact, diagonalization proves the opposite. It proves: "If you have a set of Cantor Stings that can be listed this way, then it does not include all Cantor Strings."

It is a simple trick of logic to invert the statement "If A, then B" to get the 100% equivalent statement "If NOT B, then NOT A." So diagonalization also proves "If you have a set of all Cantor Strings, then it can't be listed this way."

Elegant, no?

The reason so many Anti-Cantor cranks exists, is because we don;t seem to teach Cantor's proof correctly.

JeffJo
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    I actually make the point that you're making later on in the dialogue; look at where it says "Forget about the purported complete lists of real numbers for a moment..." The problem is, when I phrase the proof that way, the cranks' retort is that they do not think that the diagonalization procedure will always work in constructing a well-defined real number (or infinite string if you prefer). They think that the diagonalization procedure is not well-defined if you start with a list that happens to be complete. The problem is that they're misdiagnosing the source of the contradiction. – Keshav Srinivasan Apr 16 '16 at 20:27
  • I'd agree, if you you'd said "They think that the diagonalization procedure is not well-defined if you start with a list that happens to be INFINITE." The two are not the same, and I do feel it is the "infinite" part that confuses them. Not the "complete" part. – JeffJo Apr 17 '16 at 19:09
  • No, they're fine with infinite lists. Like if you start with an infinite list of the form .0111111... then .1011111... then .1101111... etc., they think diagonalization is a well-defined procedure. But they think that if the list happens to be complete, then the diagonalization process is not well-defined. That's because if the list was complete and the diagonalization procedure was well-defined, then the number being constructed would be somewhere in the list, say the Nth spot. And then Nth digit of that number would be ill-defined, because it's required to differ from itself. – Keshav Srinivasan Apr 17 '16 at 19:51
  • So they think that the diagonalization process is only well-defined for those lists which do not contain the number which would be constructed through diagonalization. Like I said, the problem is that they're misdiagnosing the source of the contradiction. They don't understand that the diagonalization process is well-defined for all lists. – Keshav Srinivasan Apr 17 '16 at 19:54
  • And I think that their difficulty comes when they try to correlate the infinite length of the strings, to the infinite length of the assumed "complete" list. The question they have - the one behind the imagined contradiction you describe - is "If the list is infinite, how can it not contain all possible strings?" The answer is that you can correlate two infinities in more than one way, and some ways miss elements of one.. – JeffJo Apr 18 '16 at 13:56
  • Like I said, they're absolutely fine with the notion that the infinite list consisting of .01111..., .10111..., .110111..., etc. is incomplete and that diagonalization is well-defined operation that generates a number not on the list in that particular case, because there's no contradiction involved in constructing that number. But in the case of an infinite list that happens to be complete, they think the diagonalization process is not well-defined because you would get a contradiction in trying to construct the number. So it's not an issue of infinite lists in general. – Keshav Srinivasan Apr 18 '16 at 14:04
  • In coming back to re-read this, I still think you missed my original point. Despite the fact that 99% of the time it is taught that way, Cantor Diagonalization is NOT a proof-by contradiction. It is a proof by contraposition. – JeffJo Sep 01 '16 at 13:08
  • Look at that "later point" you referred to. The counter-argument was "No, it's only possible to construct such a real number if that real number isn't on the list." This is still supposing that every real number is listed, an assumption Cantor NEVER makes but the CantorCranks are. The proof goes like this: Assume S is a countable infinite set of Cantor Strings. Diagonalization finds a Cantor String that is not in S. Therefore, if S is countable, then it is incomplete. Contraposition: If S is complete, then it is not countable. – JeffJo Sep 01 '16 at 13:16
  • Well, even when you don't phrase it as a proof by contradiction, their response is still "If the number being constructed through diagonalization is not on the list, then I agree that the diagonalization is well-defined and produces a well-defined real number that's not on the list. But if the list is complete, then the diagonalization process is not well-defined, because it would lead to a contradiction if it was well-defined. So I agree that for incomplete lists, diagonalization produces a well-defined number that's not on the list, but I don't think the same is true for complete lists." – Keshav Srinivasan Sep 01 '16 at 15:18
  • So they're not making the mistake of thinking that the list being complete is the only possibility, but they do think that it is A possibility, and they think that under that scenario, diagonalization doesn't produce a well-defined real number, because the number would have to differ from itself. So they think Cantor's proof just produces the tautology "Incomplete lists are incomplete." – Keshav Srinivasan Sep 01 '16 at 15:26
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First off, there are formally verified proofs of the theorem. In fact, every major theorem proof system or proof explorer has proved it, and it is number 22 in the classic "100 theorems" benchmark: http://www.cs.ru.nl/~freek/100/

The problem is one may not accept the axioms and rules of inference.

However, that is not the only problem. One may not even accept the statement as it is stated. I will show an alternative statement, which may be what the cranks have in mind.

Consider a Gödel numbering or any similar scheme from the naturals to the formulas of set theory. Clearly, this numbering lists all the possible formulas of set theory.

Since real numbers are merely formulas of set theory, the encoding will define every real number. This last statement is bound to be controversial, but nonetheless it's a valid definition of what it means to be a real number and your cranky friends could be utilizing it. (In fact, if I am understanding Asaf correctly, interestingly enough, this is the usual definition: Undefinable Real Numbers )

No idea why this got a down-vote and a delete request. The point was not brought forward in the previous answers and is based on Hamkin's answer: https://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numb#answer-44129

Simply substitute definable real number in place of real number and the diagonalization proof no longer works. Note that each real number may be definable.

Dole
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