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I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers. First, take the numbers between zero and one, exclusive. The following sequence of real numbers is suggested so that we have bijection.

0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.

Obviously, this includes repeats, but this set is countable. Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable. Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable.

Please help me with this. I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.

Thanks.

Martin Sleziak
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analysisj
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    [Liouville number](http://en.wikipedia.org/wiki/Liouville_number) is a good counter example for a non-repeating number which is not in the range of your function. – Asaf Karagila Sep 01 '11 at 13:37
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    The reals , constructed your way would be the product of ($0,1,2,...,9$) indexed by n in $\mathbb N $, which is not countable. – gary Sep 01 '11 at 13:54
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    I see this a lot with my undergrads, when did it become common to write decimal numbers between 0 and 1 without the 0 in front of the decimal point? That, and the lack of spaces after commas - I had to stare at this for a minute before I could tell what was going on. – Amit Kumar Gupta Sep 01 '11 at 18:53
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    0.2, 0.3, 0.4, ... , 0.32, 0.33, 0.34, ..., 0.332, 0.333, 0.334, etc but your set never reach 1/3 therefore it's not real numbers. –  Oct 01 '12 at 22:16

3 Answers3

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Your function ignores all the real numbers whose decimal representations are not finite, such as

$\dfrac13=0.3333\ldots$

The subset of real numbers that do have finite decimal representations is indeed countable (also because they are all rational and $\mathbb Q$ is countable).

J. M. ain't a mathematician
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hmakholm left over Monica
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  • Thank you. this makes sense. +1 – analysisj Sep 01 '11 at 13:41
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    It is a common mistake to assume that decimal recurrence is a rigorous representation in the decimal number system. It is a widely accepted convention only. The decimal system is a positional notation for representation of discrete (digital) numbers. The number of digits on the right of the decimal point determines precision. The injective relationship with the set of natural numbers mandates a non zero precision (non zero interval between numbers of a countable set) but infinite recurrence implies infinite precision (interval between numbers of zero) and this violates the injective rule. – Marcus Anderson Jun 30 '17 at 06:00
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If you know Cantor's diagonalization argument, then you should be able to find a counterexample to your problem using it. Say the method we use to make the $k$-th digit of our new real number different from the $k$-th digit of the $k$-th number in your list is adding 1 to it (and make it 0 if it's 9). Then Cantor's argument gives the number $0.21111111111111...$ Which is not in your list because (as said below) you only have real numbers with finite decimal expansions in your list.

(was previously a comment, but made an answer following suggestion)

Vhailor
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    The nice thing about mathematics is that if you have two "theorems" that contradict each other, you can usually put them head-to-head to reveal the flaw. If one says "here is an enumeration of the real numbers", and the other says "any enumeration of the real numbers must be missing one", you can put them together to find the missing one. – Tanner Swett Sep 01 '11 at 16:58
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The set you have shown is a list of all rationals between $0$ and $1$ that can be written in the form $x /10^n$ with $x\in\mathbb{Z}$, which is countable. But the full set of reals between $0$ and $1$ is bigger.

All reals are the limit of some sub-sequence of this sequence, but not all are in this sequence, e.g. $\sqrt{2}=1.14142\ldots$ or $\frac{1}{3}=0.33333\ldots$.

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    How is this different from the answers already given ***over three years ago***? – Asaf Karagila Feb 20 '15 at 14:04
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    It is different because he mentioned $\sqrt{2}. 1/3 belongs to the countable set Q. And combining the countable set Q with the set given in the question ist easy. So the real problem are irrational numbers. – user1027167 Feb 20 '18 at 10:42