'Obvious' theorems that are actually false

Question

It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true".

Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem).

But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false (i.e. counter-intuitively-false theorems)?

The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$ ).

I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$.

I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself.

Thanks!

Answer 1

Theorem (false):

One can arbitrarily rearrange the terms in a convergent series without changing its value.

Answer 2

A shape with finite volume must have finite surface area.

Answer 3

I wish I'd thought of this yesterday, when the question was fresh, because it's astounding. Suppose $A$ and $B$ are playing the following game: $A$ chooses two different numbers, via some method not known to $B$, writes them on slips of paper, and puts the slips in a hat.

$B$ draws one of the slips at random and examines its number. She then predicts whether it is the larger of the two numbers.

If $B$ simply flips a coin to decide her prediction, she will be correct half the time.

Obviously, there is no method that can do better than the coin flip.

But there is such a method, described in Thomas M. Cover “Pick the largest number”Open Problems in Communication and Computation Springer-Verlag, 1987, p152.

which I described briefly here, and in detail here.

Answer 4

This is elementary compared to most of the other examples, but how about

There are more rational numbers than there are integers.

Answer 5

I keep harping on this, because I think it's a spectacular example of something that can be demonstrated to be completely obvious (not only because it seems so, but because it was so widely believed for so long) and yet is completely wrong:

Suppose $\Phi$ is a property that might or might not hold of some object. Then there is a collection $S_\Phi$ of all objects with property $\Phi$.

Many serious and even famous mathematicians went ahead with this intuitively obvious but utterly false principle, whose demolition shook mathematics to its foundations and marks the beginning of modern logic and set theory.

(There are many counterexamples, of which the most well known is $\Phi(x) = $ “$x$ is not a member of collection $x$”. For others, see Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? $\{x:P(x)\}$ and Paradox of General Comprehension in Set Theory, other than Russell's Paradox.)

Answer 6

In a related mathOverflow thread, Gowers pointed out the following obvious but false claim:

Let $I_1, I_2, \ldots$ be subintervals of $[0,1]$ whose total length is strictly less than 1. Then the union of the $I_i$ cannot contain $\Bbb Q\cap [0,1]$.

(Note that if $\Bbb Q$ is replaced with $\Bbb R$, the claim is true.)

I find the fact that all of $\Bbb Q$ can be covered by an arbitrarily small family of intervals to be one of the most bizarrely counterintuitive in all of mathematics.

Answer 7

If a function $f(x)$ has an horizontal asymptote, then $\lim f'(x) = 0$

Answer 8

$ 0.\overline{9} < 1 $

Probably the most famous of the "obvious" but false.

Answer 9

The falseness of

Let $S$ be an infinite family of strictly positive numbers. Then $\sum S = \infty$

has been boggling people for thousands of years. It is the basis for Zeno's paradox, but if you think Zeno's paradox is old and tired, consider that it is also the basis for the Gabriel's Horn paradox (also mentioned in this thread), which still puzzles people.

Answer 10

Every chain of subsets of $\mathbb N$ is countable.

Answer 11

Keller's conjecture is obviously true:

Let $\Bbb R^n$ be completely covered with identical, non-overlapping $n$-cubes. There must be two cubes that share a face.

(For example, when $n=2$ we cover the plane with little square tiles, and the conjecture states that there must be two tiles that share an edge. This is true.)

However, the conjecture is false for all $n>7$.

Answer 12

• If $U$ is an open subset of $\mathbb{R}^n$ that is homeomorphic to $\mathbb{R}^n$, one might think it "obvious" that it's in fact diffeomorphic to $\mathbb{R}^n$ (perhaps thinking something like "topologically it looks like $\mathbb{R}^n$, and differentiably it's locally trivial"). In fact, this is true (but by no means obvious!) for $n\neq 4$. But for $n=4$ it is false: there exist exotic $\mathbb{R}^4$'s (differentiable manifolds that are homeomorphic, but not diffeomorphic, to $\mathbb{R}^4$), including "small" ones which are diffeomorphic to an open subset of $\mathbb{R}^4$.

• Much less profound, but still fun: it's "obvious" that the sum of two convex open sets in the plane whose border is $C^\infty$ also has a $C^\infty$ border (perhaps thinking something like "the border of the sum is parametrized by a smooth function of the borders of the summands"). But this is false: in fact, the border of the sum is always $C^{20/3}$ (meaning six times differentiable and with a sixth derivative which is appropriately Hölder) and no more in general. A simple counterexample is given by the epigraphs of $x^4/4$ and $x^6/6$. For details, see Kiselman, "Smoothness of Vector Sums of Plane Convex Sets", Math. Scand. 60 (1987), 239–252.

Answer 13

$i^i$ is imaginary.$\ \ \ \ \ \ $

Answer 14

This part is true (Jordan-Brouwer separation theorem):

(a) Any imbedding of the $2$-sphere into $3$-dimensional Euclidean space separates the space into two disjoint regions.

But this part, which would seem to be a natural generalization of the Jordan-Schönflies Curve Theorem, is not true:

(b) The regions are homeomorphic to the inside and outside of the unit sphere.

Answer 15

I really like "wrong proofs" as typically the insight why the proof is wrong gives you some understanding of the topic. One very simple version is this one, which I threw at my first semesters when I was a tutor:

Each binary relation which is symmetric and transitive is also reflexive and therefor an equivalence relation.

"Proof":

Let $\sim$ denote a symmetric and transitive relation and let $x$, $y$ be two elements with $x \sim y$. As $\sim$ is symmetric, it holds that $y \sim x$. Since $x \sim y$ and $y\sim x$ it follows by the transitivity of $\sim$ that $x \sim x$, which is the definition of reflexivity.

Edit: Since I was asked, here's why the proof is wrong (move your mouse there to show):

Take a look at the empty relation on a non-empty set $S$, so that there are no $x, y \in S$ so that $x \sim y$. This relation is symmetric and transitive, but it is not reflexive. Reflexivity needs $x \sim x$ to hold for all $x$. The proof assumes that there is a y so that x ~ y, which isn't necessarily the case for all $x$.

Answer 16

Theorems that are intuitively true, but actually flawed:

• There is no continuous, nowhere-differentiable real function.

• There is no real function that is differentiable and not monotonic on any non-trivial interval.

• If a real function satisfies $\forall x, y, f(x+y) =f(x) +f(y) $, it is of the form $x\to ax$.

• Infinite sums and integrals can be swapped anytime.

• A connected metric space is path-connected.

Answer 17

Here's one of my favorites: Let's assume playing with a fair coin.

Theorem (false) In a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.

The following is from W. Fellers classic of Introduction to Probability Theory and It's Applications, Vol 1:

According to widespread beliefs a so-called law of averages should ensure the Theorem above. But, in fact this theorem is wrong and contrary to the usual belief the following holds:

With probability $\frac{1}{2}$ no equalization occurred in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.

In fact this leads to the Arc sine law for last visits (see e.g. Vol 1, ch.3, section 4, Theorem 1).

Note: Please note the remarkable statements cited from Chapter III: Fluctuations in Coin Tossing and Random Walks:

For example, in various applications it is assumed, that observations on an individual coin-tossing game during a long time interval will yield the same statistical characteristics as the observation of the results of a huge number of independent games at one given instant. This is not so.

and later on:

Anyhow, it stands to reason that if even the simple coin-tossing game leads to paradoxical results that contradict our intuition, the latter cannot serve as a reliable guide in more complicated situations.

[2015-07-16] According to a comment from @HenningMakholm some examples exposing striking aspects.

• Suppose that a great many coin-tossing games are conducted simultaneously at the rate of one per second, day and night, for a whole year. On the average, in one out of ten games the last equalization will occur before $9$ days have passed, and the lead will not change during the following 356 days. In one out of twenty cases the last equalization takes place within $2\frac{1}{2}$ days, and in one out of a hundred cases it occurs within the first $2$ hours and $10$ minutes.

• Suppose that in a learning experiment lasting one year a child was consistently lagging except, perhaps, during the initial week. Another child was consistently ahead except, perhaps, during the last week. Would the two children be judged equal? Yet, let a group of $11$ children be exposed to a similar learning experiment involving no intelligence but only chance. One among the $11$ would appear as leader for all but one week, another as laggard for all but one week.

The examples above are in fact a consequence of the Arc sine law for last visits.

Answer 18

The following statement I once believed to be "obvious":

If $f:\mathbb{R} \rightarrow [0,\infty)$ continuous is such that $\int_{-\infty }^{\infty }f(x)\text{d}x<{\infty } $, then $\lim \limits_{x \to \pm\infty} f(x) = 0$

which is actually false.

(Note: It is true if $f$ is uniformly continuous!)

Answer 19

There are a good number of counterintuitive probability situations out there. One of my favorites is nontransitive dice:

There are 3 dice, A, B and C. The dice have numbers from 1-9 on their sides (repeats possible). If die B beats (higher number) die A more than half the time and die C beats die B more than half the time, then die C will beat die A more than half the time.

This is not necessarily a true statement. Dice can be designed such that the "x beats y" property is not transitive. A beats B, which beats C, which beats A.

Answer 20

The real numbers/Cantor set are countable.

There are several false "obvious" proofs:

1. "Proof". Consider the tree $\{0,\ldots,9\}^\Bbb N$, then every real number corresponds to a node in the tree. Since there are only countably many levels and each is finite, it follows that the real numbers are finite.

   Why does it fail? This set is actually not a tree. You can order it so it looks like a tree, but in fact the tree would be composed of initial segments of each functions ordered by continuation. This tree, then, would have a last level (namely a level that no point there has a successor), and it would be exactly the level of the functions themselves (the previous levels would be proper initial segments of the functions).

   If we remove that last level, then the tree is indeed countable, but now each real number corresponds to a branch in the tree rather than a node. (It's the unique branch whose limit equals to the function, which previously appeared on that final level.)

2. "Proof". The rational numbers are countable, and between every two real numbers there is a rational number. Therefore this defines a bijection between pairs of real numbers and the rational numbers.

   Why does it fail? Because there are many, many, many pairs being mapped to the same rational number, this is not actually a bijection.

3. "Proof". The Cantor set is closed, its complement is open, so it is a countable union of intervals, so the Cantor set is countable.

   Why does it fail? Because not every point in the Cantor set is an endpoint of such interval. For example $\frac14$. It is true that the endpoints of these intervals form a countable dense subset, though.

4. BONUS!, $\mathcal P(\Bbb N)$ is countable.

   "Proof". For every finite $n$, $\mathcal P(n)$ is finite, and $\mathcal P(\Bbb N)=\mathcal P(\bigcup n)=\bigcup\mathcal P(n)$, is a countable union of finite sets, which is countable.

   Why does it fail? Because the union only includes finite subsets of $\Bbb N$, but none of its infinite subsets.

Answer 21

Hypothesis: Every infinitely-differentiable function is real-analytic somewhere.

This is false, as shown by (for example) the Fabius function.

Answer 22

I'm surprised noone gave this answer already, so here it is:

There are more integers than there are natural numbers.

It's obvious, isn't it?

Answer 23

Image of a measure zero set under a continuous map has measure zero!

Answer 24

The probability that you hit any single point on a dart board is $0$ but the probability that you hit the dart board is $1$ (as long as you're not as bad as I am at throwing darts ;D).

EDIT:

As @JpM pointed out I didn't follow the format of these posts albeit the idea can (easily in my opinion) be understood from what I've said above.

Pseudo-Claim: The probability of hitting a single point on a dart board is greater than $0$ since the probability of hitting it at all (assuming that you will hit the dart board) is $1$.

Seems obvious in the sense that a bunch of $0$ can't add up to be $1$ so each point must have some probability. Actually false because of some properties of measures.

Answer 25

My "theorem":

The statement Everybody loves my baby, but my baby loves nobody but me is about a pair of lovers

It is so simple and so obvious, even my grandma will understand it. And no matter how much you explain the simple logic calculation which shows that we are talking about a single narcissist here, half the class of first-semester logic students will continue insisting that your proof is wrong, and they don't know what is wrong about it, but it cannot refer to a single person.

Answer 26

Consider a function $f:(0, \infty) \rightarrow \mathbb{R}$ that is $\mathcal{C}^\infty$ on that interval. At first glance, one might think that, if $\lim(f) = 0$ as $x \rightarrow \infty$, then $\lim(f') = 0$ as $x \rightarrow \infty$. However, this is false. Here is but one counterexample:

$$f(x) = \frac{1}{x}\sin(x^2)$$

Further, if we add the stipulation that $f$ also be monotonic, counterexamples can still be found (though they are quite pathological).

Answer 27

A simple arc (homeomorphic image of the closed unit interval) in the plane has $2$-dimensional Lebesgue measure zero.

Answer 28

Any real number can be computed somehow.

More formally:

For every real number, there exists a finite-length program that computes that number.

Since real numbers are uncountable while computable numbers are countable, that just can't be the case.

This limitation comes from the fact that we're stuck using finite-length programs. Infinite-length programs can be defined to compute any real number (trivially). So there is a sense in which all real numbers can be computed.

Just not by humans. Note that, since a single infinite-length program would take up infinite memory (and we don't seem to have any infinite computers/brains), the majority of these infinite-length programs can never be known, let alone computed. So computable numbers are only those numbers computable by a finite-length program. And the set of finite-length programs is countable.

Answer 29

Theorem: Let $f_1(x,y)$ and $f_2(x,y)$ be two joint probability densities, each having its $x,y$ components positively correlated ($Cov_1(x,y)>0$, $Cov_2(x,y)>0$). Let $f_3=\alpha f_1 + (1-\alpha) f_2$ be the mixing density, for some $0\le \alpha\le 1$. Then $Cov_3(x,y)>0$.

In words: mixing populations preserves the correlation sign. In other words: if the average MSE male user is brighter than the mean, and if the average MSE female user is brighter than the mean, then the average MSE user is brigther than the mean. Obviously true.

False. See Simpson's paradox.

Answer 30

Claim:

If the dot product of two vectors is 0, then they are linearly independent.

My prof threw this question at me today and I fell for it.

Answer 31

In my opinion, the most interesting (but also sometimes not intuitive) results in mathematics are those that state a theorem that ends up being false because it actually holds in many cases, except for very few or very strange cases. In other words, the most "obvious" false theorems to me are those that have very difficult counterexamples.

Some examples:

• Banach-Tarski: There exists a strict subset $A$ of the Euclidean $n$-ball $B$ such that one can partition $A$ and $B$ into an equal number of further subsets that can be mapped to each other by isometries. This shows that not all sets are measurable, and that it's possible to perform partitions that do not preserve measure.

• Non-finiteness of differentiable structures: For $\mathbb{R}^n$ with $n = 4$, there are an uncountable number of distinct differentiable structures.

• Divergence of Fourier series: There exists an integrable function on $[-\pi, \pi]$ whose Fourier series diverges everywhere. This is extremely unusual because for any typical function we might write down, usually its Fourier series might diverge at one or a finite number of points, but will probably converge everywhere else.

Answer 32

An 'obvious' but false theorem: There are more open sets in $\mathbb R^2$ (or $\mathbb R^n$) than there are real numbers.

And in a similar vein we have this corollary to the first statement: There are more continuous functions $\mathbb R\rightarrow\mathbb R$ than there are real numbers.

(Both statements are false.)

Answer 33

Here are some of the false statements popping into my mind that made me raise at least one eyebrow when I first realized they were not true.

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Every linear function between two vector spaces is continuous.

True only as long as the domain is finite-dimensional. If it is not, then there exists a linear function that is not continuous—at any point!

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The set of real numbers can in no way be (totally) ordered in such a way that every non-empty set in it has a least element.

False if choice is assumed, by the well-ordering theorem.

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$\mathbb Q$ is not countable.

I am still tempted to believe it sometimes...

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If the derivative of a continuous real-to-real function exists almost everywhere and (wherever it exists) vanishes almost everywhere, then the function must be constant.

False. In fact, there exists a function that satisfies the premise and it is strictly [sic!] increasing!

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Any compact set is closed.

The name “compact” would suggest this, but this can be guaranteed only in Hausdorff spaces.

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A set is compact if and only if every sequence in it contains a convergent subsequence.

While true in metric spaces, not only is it false in some more general topological spaces, but also neither condition implies the other!

Answer 34

The Hauptvermutung states that there is essentially only one PL structure on a manifold. More precisely, it states that any two triangulations have a common subdivision. The reason why this seems "obviously true" is that you can take both triangulations and superimpose them one on top of the other, subdividing the manifold into a bunch of cells, and then taking the barycentric subdivision to get a triangulation. It turns out that this is false and one needs some pretty subtle invariants to detect it. The problem with the argument that I gave is that one triangulation could be very wild with respect to the other (fractally wiggly) so that their union does not subdivide the manifold into a nice collection of cells.

Answer 35

Stein's paradox is to me the most puzzling mathematical notion I've ever known (although I'm not a mathematician), mostly because it's not a mathematical "artifact", but its non-intuitiveness carries very tangible error consequences.

Theorem: (false)

One can do no better than ordinary decision rule for estimating the mean of a multivariate Gaussian distribution under mean squared error.

In other words, completely independent phenomena can actually be combined for a lower joint estimation error.

Answer 36

What about this:

$\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic (as Abelian groups with addition).

It falls under the category of "Let's take the Hamel basis of $\mathbb{R}$...", but I like it a lot.

Answer 37

I think this is not covered in any of the other answers (although, to be sure, there are a lot of them). The Simpson's paradox one is close, but I think this is different and somewhat easier to understand:

If $X$ is positively correlated with $Y$, and $Y$ is positively correlated with $Z$, then $X$ is positively correlated with $Z$.

In other words, positive correlation is transitive. I think it's fairly intuitive, yet false.

Answer 38

"Obviously" $$(x^y)^z = x^{y\cdot z}$$ for $x,y,z \in \mathbb{C}$ such that given expressions are defined.

Answer 39

"A sequence of numbers in which every number is larger than the previous, will always eventually go above a given value L."

Answer 40

I really want the following to be true:

Theorem: Let $S$ a subset of a vector space. If $S$ is pairwise linearly independent (meaning each $\{v,w \} \subseteq S$ is linearly independent) then $S$ is linearly independent.

And yet, it is false. For example, $$ \{ v,w,v+w \} $$ If $S$ only had two elements then we win by default. In any event, students tend to believe this. I mean, it's linear algebra, the principle of superposition ought to apply right? Something is the sum of its parts, linear independence begets linear independence... very seductive, very wrong.

Answer 41

Something I used to be seduced by in my mathematical immaturity (which is sadly still existing):

Suppose that $P_n$ are a family of statements indexed by $n\in\mathbb{N}$ and we can assign meaning to $P_{\infty}$. Then if $P_n$ is true for all $n\in\mathbb{N}$, then $P_{\infty}$ is true also.

Answer 42

One of the first times I got caught out being wrong about something so obvious was believing:

abs(x) is never equal to -x

Of course abs(x) is defined as -x for x < 0

Answer 43

Cauchy's theorem implies that:

if one makes a physical model of a convex polyhedron by connecting together rigid plates for each of the polyhedron faces with flexible hinges along the polyhedron edges, then this ensemble of plates and hinges will necessarily form a rigid structure.

However, there are counter-examples if you allow a general polyhedron (not convex).

Answer 44

Before 1955 everyone “knew” that to know the nth decimal digit of $\pi$ (and for any other irrational) it was necessary to know the previous digits. A genius like Archimedes ("There was more imagination in the head of Archimedes than in that of Homer": Voltaire) "knew" very well this as History shows. However, The Bailey–Borwein–Plouffe formula (BBP formula) finished with this sacred for centuries “knowledge” and now it is possible to know, for example, the 33-th digit without to know the precedent ones.

Concerning the intuitive perception, it is false that a continuous numerical function must be derivable at least in one point; it is false too that a little square cannot contains a curve of infinite length.

Answer 45

EDIT: The counterexample I had in mind is incorrect; I've asked another question to try and clarify the matter one way or the other. Regardless, I think it's safe to say that this is no longer obvious! But I'll go ahead and update (or delete) my answer accordingly once I've gotten a bit more clarity.

Here's a topological example that takes some thought to falsify: roughly, 'for every non-intersecting curve between two opposite corners of a square, there's a curve between the other two corners that only intersects it once'. Formally:

Let $f: [0, 1]\mapsto [0,1]^2$ be a non-self-intersecting curve with $f(0) = (0,0)$, $f(1) = (1,1)$, and $f(t)\in (0,1)^2$ for $t\in(0,1)$. Then there exists a non-self-intersecting curve $g: [0, 1]\mapsto [0,1]^2$ with $g(0) = (1,0)$, $g(1) = (0,1)$, and $g(t)\in (0,1)^2$ for $t\in(0,1)$ such that there are unique $t_0$ and $ t_1$ with $f(t_0) = g(t_1)$.

This seems obvious (at least to me) on first glance, and even on second glance, the example of the Jordan Curve theorem suggests that it should be true; after all, we get a 'left side' and a 'right side' of our curve by the JCT, and doesn't the Schoenflies theorem mean that we should be able to find an inverse mapping of our curve to the circle? But it's false; there are curves $f()$ that can't be intersected only once by any curve $g()$. Finding a counterexample makes a nice exercise...

Answer 46

The following is obviously false, but it is actually true, as shown in the Wikipedia article about Vitali sets.

The exists a countable collection $\left\{V_n\right\}$ of subsets of the unit circle such that:

1. Any two distinct $V_n$ are disjoint.
2. Any $V_n$ can be obtained from any other by a rotation.
3. The union of all the $V_n$ is the whole circle.

All the $V_n$ must have, from property two, the same "size" (for any reasonable definition of "size"), but if the above fact was true, the sum of their (equal) sizes would be the size of the circle (positive, but finite). But if the size was zero, the the sum should be zero, and if the size was positive, the sum should be infinite.

A consequence of this is that the following is false (although we all would like it to be true):

There exists a function $\mu$ that, given a bounded subset of $\mathbb{R}$, tells you its "size". Precisely:

1. If $A\subset\mathbb{R}$ is bounded, then $\mu\left(A\right) \in \left[0,\infty\right[$.
2. If $\left\{A_n\right\}_{n\in\mathbb{N}}$ is a sequence of bounded disjoint subsets of $\mathbb{R}$ (that is, $A_n \cap A_m=\emptyset$ whenever $n\neq m$) with bounded union (that is, $\bigcup A_n$ is bounded), then $\sum\mu\left(A_n\right)=\mu\left(\bigcup A_n\right)$.
3. If $A$ is bounded, $x$ is a real number, and we define $A+x=\left\{a+x:a\in A\right\}$, then $\mu\left(A+x\right) = \mu\left(A\right)$.
4. $\mu\left(\left[0,1\right]\right) = 1$

Indeed, rewriting the first fact exchanging the circle by the half-open interval $\left[0,1\right[$ and exchanging rotations for cyclic shifts "mod $1$", we realize that if $\mu$ satisfies the first three conditions above, then $\mu\left(A\right)=0$ for all $A$.

Answer 47

The Birthday Paradox

If 30 people are randomly selected, and they have birthdays that are independently, (identically) uniformly distributed over the calendar year, then the probability that two (or more) of them have the same birthday is approximately $\frac{1}{12}$.

Answer 48

A linear order can be uniquely (up to isomorphism) reconstructed from the set of order types of its proper initial segments.

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Update: Even if we know the cardinality of the linear order, and know that it does not have a maximal element, this "theorem" still does not hold.

Answer 49

All infinities are of the same size.

But Cantor's theorem shows otherwise.

Answer 50

Here are some "obvious" statements to which Richard's paradox can apply to:

1. For a given predicate $P$, there exists a set $S$ of $x$ for which $P(x)$ is true. (Russell's paradox)
2. The set of integers and the set of real numbers are the same infinite size. (Cantor's diagonal argument.)
3. There exists a formalization of arithmetic in which all true statements are exactly those which are provable. (Gödel's theorem(s))
4. There exists a computer program (Turing machine) that can effectively determine if any other computer program doesn't halt. (Halting problem)

Answer 51

There are a lot of examples in (extremal) graph theory, where an obvious argument shows that a statement is true, except that there are a number of small counterexamples which are easy to overlook.

Consider the following statement: Let $G$ be a graph with $n$ vertices and the largest number of edges subject to the condition that $G$ does not contain a pair of disjoint edges (i.e. $K_2 + K_2$). Then $G$ is a star (i.e. $K_{1,n-1}$).

This is obviously true, if you think about it for a moment. But for $n=3$, a better solution is to take $G = C_3$. And for $n=4$, taking $C_3$ plus an isolated vertex is just as good as taking $K_{1,3}$.

Answer 52

A subgroup of a finitely generated group may not be finitely generated and there are up to isomorphism at most two $pq$ groups, where $p$ and $q$ are prime.

To strengthen your example, $\sum_{p \ prime} 1/p$ diverges.

The continuum hypothesis also seems like it has to have in answer in ZFC, which it doesn't.

On another page, mathematicians thought that cyclotomic fields "obviously satisfy the unique factorization theorem" leading to some false proof attempts to Fermat's last theorem.

Next, one might think that the "angle trisection" is possible or that any set of analytic functions $\{f_\alpha\}$, such that for every $z \in \mathbb C$, the set $\{f_\alpha(z)\}$ is countable, itself has to be countable.

These are just some random examples that came to mind and since the term "obvious" is subjective, you may very well disagree with items on my list. I guess it heavily depends on your mathematical background.

Answer 53

The following is a very well-known example, though probably slightly outside of the world of mathematics, rather physics. A great many people would 'intuitively' consider the following to be true:

The heavier the object, the faster it falls down.

In fact, the story goes that this was supposed to be common knowledge until Galileo Galilei disproved it (as the story goes, by dropping two balls of the tower in Pisa, which never happened though).

One of the first physics classes many people have (I'm talking primary school here) aims at showing this theorem is false, and actually everything falls with the same acceleration (ignoring resistance by air) regardless of weight.

Answer 54

If you start thinking about the rigidity of thin shells in $\mathbb{R}^3$, you quickly encounter a slew of counterintuitive results.

For instance, it is obvious that a spherical shell is ($C^2$) rigid, and this is in fact true. A smooth, closed, compact surface with everywhere positive Gaussian curvature is likewise rigid. One might imagine these results generalize to

• Any closed surface;
• Any closed surface with positive Gaussian curvature everywhere but at finitely many points;
• Any surface with boundary with everywhere positive Gaussian curvature;

and all of these are false.

Moreover, after thinking about reflections, or poking and prodding at a ping-pong ball, it is intuitively obvious that a spherical shell is not $C^0$ rigid. But you can't really "see" any difference between a $C^1$ and a $C^2$ deformation of the sphere, so surely the sphere is $C^1$ rigid? Far from it -- given any arbitrary closed surface that is topologically a sphere, and distance $\epsilon$, it is possible to $C^1$-isometrically embed a sphere $\epsilon$-close to the target surface!

Answer 55

Here is a proposition, when put in layman's terms, an average person would claim to be true.

Every subset of real numbers has a measure.

How can this be false, when you mark a region, say in two dimensions, of course, it has an area?! Unless you constructed a Vitali set at some point, we tend to think that the concept of length/area/volume should extend to all possible subsets.

Here is another such false proposition.

Axiom of Determinacy

If we are playing a two-player infinite game where we create a real number in $[0,1]$ by choosing decimal digits in turns and one of us tries to land the resulting number in a pre-determined payoff set that is known to both of us and the other tries to avoid it, how could it be that there is a game where neither of us have a winning strategy? We both have complete information about the payoff set what numbers to avoid and what numbers to hit, one of us should be able to come up with a strategy. Well, unfortunately no.

Both of these propositions are inconsistent with the axiom of choice, with which you can construct the counterexamples that will not "nicely behave".

Fact: The latter proposition implies the former. (in ZF, with which AD is believed to be consistent).

Answer 56

It's not exactly a theorem, but it fools every math newcomer:

$e = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n$

$(1 + 1/\infty)$ is $1$, obviously. And 1 to the power of $\infty$ is obviously still 1.

Nope, it's 2.718...

Answer 57

$\textbf{Counter-Intuitive Example}$

$$\ \ \textbf{D}_v f(\textbf{a}) = 0, \forall \textbf{v},a \not \Rightarrow f \ \ \text{continuous}.$$

Answer 58

Geometry proofs done informally by drawing figures on the blackboard. You then bypass the axioms of Euclidian geometry, you pretend that you don't need to invoke them as the figures drawn seem sufficient. However, in Earth's gravity Euclidean geometry is only an approximation.

Answer 59

Infinitely many terms always have a sum equal to infinity.

Answer 60

Another example of 'obvious' being not true is Bus Waiting Time Paradox.

If the mean time between two consecutive buses arriving to a bus station is $M$, one should expect that the mean time you have to be waiting in the station before the next bus arrives is also M. But this is not true; and depending of the particular bus arrival time distribution you'll have to wait a time $M'\geqslant M$

Answer 61

Someone else mentioned "there are more rational numbers than integers". Along the same lines, I had a hard time accepting that

There are more integers than there are real numbers between 0 and 1

is false. I mean, I get it now, but it seemed intuitively very wrong to me before I studied transfinite numbers.

Answer 62

Perhaps this isn't quite what you were looking for, but it's still fun! How about a proof that is obviously incorrect but (to newcomers) it is difficult to figure out what is wrong.

Let $x = y$. Then $$x^2 = xy$$ $$x^2-y^2=xy-y^2$$ $$(x+y)(x-y)=y(x-y)$$ $$x+y = y$$ $$2y=y$$ $$2=1$$

Answer 63

An "obviously true" theorem:

If take a 3d object $U$ and chop it into finitely many pieces, any object $V$ I rearrange those pieces into will have the same volume as the object $U$ I started with.

But in fact, the Banach-Tarski paradox tells us this isn't true -- if we construct our finitely-many subsets of $U$ "weirdly" enough, we can actually build a $V$ with any volume we'd like.

Answer 64

Any continuous function is differentiable at least somewhere, right?

False, the Wierstrass function is a famous counterexample

https://en.wikipedia.org/wiki/Weierstrass_function

It is continous everywhere, yet it is differentiable nowhere.

Answer 65

Here is one thing commonly thought to be true but is quite horribly wrong on many accounts:

There is a notion of mathematics where we can say things are 
"actually true" or "actually false".  

An example of making this error: the OP. Other examples: the many responses.

There are several reasons why this is wrong. First, in the system most mathematicians assume when not being explicit, we have no standard model (we have no models within that system because any model would show the system consistent which we know we cannot show in that system through blah blah Godel blah.. I know you don't want details, just explaining what I'm getting at). Truth and falsity are semantic - they exist in models and so without one, we don't make claims of truth or falsity.

But also, mathematics is not "the system most mathematicians assume when not being explicit" - it is formalisation in general. There are many systems seriously investigated by mathematicians that make numerous "counterintuitive" derivations. For instance, these are all obvious and wrong in different systems:

• A statement cannot be both true and false. (In paraconsistent logics, statements can be both true and false and the system does not collapse to trivial - in fact, a number of dialetheists argue this is a much more accurate logical system for real world reasoning).
• There are discontinuous total functions. (In a number of constructive systems it is not possible to prove the existence of discontinuous total functions. Some are even strong enough to prove that all total functions are continuous.)
• Every infinite set A has the same cardinality as AxA. (This is not necessarily true in systems without the axiom of choice. Famously, Tarski tried to publish his result on this implication and was rejected by both Frechet and Lebesgue. Frechet thought the paper was obvious and well-known and had no mathematical merit. Lebesgue thought both the axiom of choice and the implication from it were both wrong, so the paper had no mathematical merit.)

I only bring these examples up not as answers to the OP but simply to illustrate my real answer that the question itself demonstrates an extremely common assumption in mathematics that is in fact wrong.

EDIT

This is an area that I think is often a place of common misunderstanding, and discussion in the comments makes it clear I should elaborate. Modern mathematics separates out the domains we make statements on into syntax and semantics.

Syntax

The syntax is the theory - the formal language, axioms specified as sentences in the formal language, and some metalogical rules of inference. In the syntax, we talk about sentences, propositions, terms, derivations, and proofs. It is a place of symbol manipulation.

Semantics

The semantics is the model - it is the meaning we ascribe to the statements of the theory. An interpretation of a theory is a model that assigns to each formula of the theory a meaning value - typically truth. Truth is semantic and is specific to a model.

The "problem"

A model is a consistent interpretation of the truth meaning of a theory. If a theory has a model, it has almost trivially been shown to be consistent. But... it is well known that a theory strong enough to express the Gödel diagonalisation can never prove it's own consistency. For these theories, we will never have a model and cannot make statements about the meaning of any formula.

In these theories, it is wrong to talk about truth or falsity. We don't have a model giving meaning to that. We will never have a model.

That's not really a problem. For centuries, mathematicians had loosely combined derivation and truth and had mostly discussed them as one thing. Derivation and proof were seen as the important part of mathematics and formalization. You still have that.

Also, it is perfectly meaningful to derive results that say "if this theory is consistent and has a model, then...". Model theory has been doing that for nearly a century.

What about truth predicates?

But people seem to want more. They want to talk about truth, as that is a form of meaning that holds a special place. They often go to great lengths to try to continue to assign truth and falsity. One common approach is to form truth predicates - predicates in the syntax that have the property that asserting the predicate on a formula corresponds to asserting the validity of the statement (that it is true in all models).

Note the switch - a truth predicate is syntactical. We still aren't talking about true or false here - the context of their use is still whether statements including the predicate "are derivable" or "obtain". Theories may have multiple models - most theories are not categorical just from things like Löwenheim–Skolem, so predicates cannot talk about truth. They can talk about validity - and that's really what is going on here - but even that is extremely problematic.

Incomplete theories cannot actually derive anything about validity on the total theory. And actually, this is where Tarski's theorem on nondefinability comes in and it is shown that such a predicate doesn't actually exist. So others keep at it with a hierarchy and reflection extensions of the base theory, seeking out some approximation of a fixed point for validity.

But this doesn't actually buy anything to do with insight into truth. It cannot. There is nothing you can do to reach truth because you cannot know if the theory is consistent or not and whether truth exists. And no attempts to reach beyond derivability actually give a predicate that can be used and say "this is true". The predicate is only useful to say "this is provable".

But there are already provability predicates, and that investigation is much more profitable. Truth predicates are a voiceless oracles. They do not help anyone make assertions on truth. They are simply reformulations of "if we knew that X was consistent, and we had some platonic sight that could see the truth values in all models, and we could collate the infinite possibilities and see the validities forever hidden, then this predicate applied to this class of statements would agree with those assertions that are valid". But if we had that supernatural sight, we could more easily just say "hey, that's true in that model - and that's false over there." Without that, we can use the predicate to say "truth is preserved in this derivation". Which doesn't add anything.

A truth predicate doesn't talk about truth. It is irrelevant to the point.

So...

So.. life goes on. My whole point in posting this answer was to illustrate that the initial question was making a common obvious assumption that is actually wrong. You should not talk about truth in the commonly used ambient theory - just talk about what is provable and you are fine. If you want to talk about truth, ensure you specify the ambient theory and it is one where such discussions are meaningful. Or talk about conditional models as model theorists do.

It may not be intellectually satisfying to some people. Clearly, as of writing this, my answer has received 3 downvotes and two upvotes, so it doesn't sit right with some anonymous readers of a math web site. But there is nothing controversial about the point. It has been known for almost 100 years and it is still a common mistake.

Answer 66

For me a nice example of all of "evidence" suggesting it was true is

$$\pi(x) < \operatorname{li}(x)$$

until Skewes showed that $\pi(x) - \operatorname{li}(x)$ changes sign infinitely often

Answer 67

The following statement is wrong:

The inner angles of a triangle always sum to 180 degrees.

While it sounds plausible that the sum of the angles is a constant, it is actually a property of the space. In Euclidean space the inner angles of a triangle always sum to 180 degrees.

Answer 68

If a propositional calculus A contains all theorems of propositional calculus B under detachment and uniform substitution for propositional variables, but B does not contain all of the theorems of A, then one of the shortest single axioms of A is longer than any of the shortest single axioms of B. Or one might more haphazardly say "if propositional calculus A is bigger than propositional calculus B, then one of the shortest single axioms of A is longer than any of the shortest single axioms of B."

Answer 69

An analytic function with compact support vanishes identically.

Answer 70

Lebesgue once stated that the projections of Borel sets in $\mathbb R ^2$ on to one of its axes are also Borel sets. This fact is actually false, the realization of which is attributed to the short-lived mathematician Mikhail Yakovlevich Suslin.

Unfortunately it is very difficult to find a counterexample. The only one I've ever seen takes a result in descriptive set theory about $\mathbb N ^ {\mathbb N}$ and uses the fact that it's homeomorphic to $\mathbb R$.