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You just have to love Cantor's diagonal argument. Its simple description is easy to follow, but the results are so unexpected. I equate my astonishment of the results to the catchphrase: No Way! Way!

I have "yet again" devised a set of numbers that brings up questions.

Consider the set of all the infinite natural numbers. These are numbers like 1111... and 1212..., who are elements in the set of natural numbers. Now, do the diagonal argument on them and they turn out to be uncountable, too. But wait, the set of infinite natural numbers is a subset of natural numbers. So, the natural numbers are uncountable. Oops.

Is there a problem with my assumptions? I think 1111... should be a natural number. If so, then the set of all infinite natural numbers should be valid.

The place where I see problems is that mystical boundary between finite and infinite elements. It seems like finite natural numbers are used for the counting. Since the number is always finite, you can never reach infinity. Like a limit, you can get close, but never actually get there. Infinity wins.

I have read elsewhere, that theorems on sets can get tricky (my wording) when trying to extend a theorem from the finite case to the infinite case. But I don't know the details of the issues.

Xander Henderson
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Rick R
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  • Any set which has a bijection with $\mathbb{N}$ is called countably infinite. – Ak3.14 Feb 06 '25 at 05:10
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    1111... and 1212... are not natural numbers. There are no “infinite natural numbers.” – Martin R Feb 06 '25 at 05:27
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    "Is there a problem with my assumptions? I think 1111.... should be a natural number." Yes that is a problem with your assumptions. – spaceisdarkgreen Feb 06 '25 at 05:27
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    You think $1111\dots$ should be a natural number. Fortunately, Mathematics doesn't care what you think. – Gerry Myerson Feb 06 '25 at 05:40
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    Every nonzero natural number has a unique predecessor. What is the unique predecessor of $111111...$? – 5xum Feb 06 '25 at 06:55
  • There are no infinite natural numbers, but take a look at https://en.wikipedia.org/wiki/Ordinal_number – PM 2Ring Feb 06 '25 at 06:58
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    Theoretically, I can count to $15,$ and I can count to $3884,$ and I can count to $111 \ldots 1$ $(875$ digits), and I can count to ${10}^{{10}^{10}}$ (a number with $10^{10} + 1$ digits), but I can't count to $111\ldots$, can you? How many digits does $111 \ldots$ have?? – Dave L. Renfro Feb 06 '25 at 08:33

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I know of no reasonable interpretation of extending infinitely many digits to the right from a "natural" number, unless of course you have a decimal point, in which case you're talking about real numbers. Expressions like "$1111\ldots$" or "$1212\ldots$" don't really mean much.

To get a bit closer to your intuitions, there is a reasonable interpretation of extending infinitely many digits to the left. In particular, it's reasonable to write $\ldots1111$ and $\ldots1212$ to represent particular $10$-adic integers.

Now, relevantly to your question:

  1. The ring of $10$-adic integers is uncountable, and the diagonal argument proves it.
  2. The set of natural numbers is a subset of the ring of $10$-adic integers, not the other way around!
  3. The set of natural numbers is countable. There is no diagonal argument that applies to it.
  4. There is no contradiction among (1), (2), and (3).
Chris Culter
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    Some other issues with $10$-adic numbers (or other $n$-adic numbers): If $n$ has two or more prime divisors (such as $10 =2\cdot 5$), the $n$-adics have zero-divisors (numbers $a \ne 0, b\ne 0$ with $ab = 0$). There is no ordering on the $n$-adics (for any $n$). In general, you cannot $a < b$ or $b < a$. The concept is not defined, nor is it possible to make such a definition that is compatible with addition and multiplication. And the reason we call them the "$n$-adics" is that each $n$ (other than powers of primes) gives a different number system. – Paul Sinclair Feb 07 '25 at 19:02
  • When I selected the infinite natural numbers (INN), I wasn't worried about any use for them. I just wanted them to exist and collect them in some set. The 10-adic integers are very close to the idea of INN I was looking for. – Rick R Feb 08 '25 at 23:06
  • If the 10-adic integers are uncountable, then the positive subset is also uncountable. If not, the 10-adic integers could be constructed from 3 countable sets. It looks like an identity function between the natural numbers and positive 10-adic integers exists and is a bijection. What am I missing here? – Rick R Feb 08 '25 at 23:24
  • @RickR you'll want to start with PaulSinclair's comment, explaining some of the ways the 10-adic integers differ from the ordinary integers. If you want to know more, it would probably be better to ask separate questions (https://math.stackexchange.com/questions/ask) than to continue in the comments here. But very briefly... every natural number has only finitely many nonzero digits. By contrast, a typical 10-adic integer has infinitely many nonzero digits. That should already paint a vivid picture for why there is no bijection of the kind you're looking for. – Chris Culter Feb 09 '25 at 04:08