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Obviously the proof is flawed in some way, but I have been struggling to find out where the flaw is. I am self taught and fairly new to set theory so the flaw may be obvious (sorry):

The idea is that we can form a bijection from the Natural numbers to the Reals by organising the Reals into a set this way:

We start with 0, then we add 1 and -1 as well as the first "level" of decimals for 0 (0.1 to 0.9). Then -2,2 the first "level" of deicimals for -1 (-1.1 to -1.9) and 1 (1.1 to 1.9) and the second "level" of decimals for 0 (0.01 to 0.99). Then -3 and 3, 1st "level" for -2 and 2, 2nd "level" for -1 and 1, 3rd "level" for 0 (0.001 to 0.999)

If we keep doing this and we will be able to list every real in such a way that we can form a bijection with the natural numbers.

I feel like I did a poor job of explaining what I mean. Here's an image that I made that will (hopefully) make things look a lot clearer:link.

We will end up with a sequence starting with this: 0,1,-1, 0.1-0.9 ,-2,2, -1.1 to -1.9, 1.1-1.9, 0.01-0.99, -3,3...

Can someone explain why this doesn't work? Thanks in advance.

Asaf Karagila
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    Probably uncountably many other duplicates as well. About a third of the results of searching "real numbers countable" are probably related. – Asaf Karagila Jun 19 '18 at 10:35

2 Answers2

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All you can get from such a list are the decimal numbers. For instance, since $\frac19=0.11111111\ldots$, it will never appear there.

José Carlos Santos
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There is discussion of a very similar thing here. Indeed, recurring decimals such as $\frac{1}{3} = 0.333 \cdots$ are omitted from the list you construct.

mathphys
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    "Very similar thing here", where "here" is a link to another question on this site, means that this is a duplicate. Flag, vote to close, yes. Post a comment as an answer, no. – Asaf Karagila Jun 19 '18 at 10:31