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I have difficulties with Cantors proof. The main concern is that proof generates a “new” number which doesn’t have a corresponding natural number. So my question is split up into two questions:

  1. If we generate a number, doesn’t it have a corresponding N by definition, because we have already said that these two sets are bijective?
  2. If we use “mirror” bijection rule $(0:1)$ so we get $1$ out of $0.1$ and $12345$ out of $0.54321$, wouldn’t that imply that if there is a Cantors new number, there is already a unique natural corresponding number?
Dang Dang
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    There have been many many many previous questions about Cantor's theorem on this site. Please use the search function and see if any of them answer your questions before asking a new one. With respect to the specific questions you've asked here: (1) do you understand how proof by contradiction works? this is the contradiction. (2) Where does the "mirror bijection" map $\frac{1}{3}$? – Alex Kruckman May 03 '25 at 12:09

1 Answers1

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The diagonal argument is often presented in a confusing way that makes in harder to understand for newcomers than it ought to be.

In particular, it is common to start by saying something like "assume $f$ is a bijection $\mathbb N\to\mathbb R$", but that is pretty misleading, since the argument doesn't really depend on the "bijection" assumption.

The right way to frame the argument is:

Assume that $f$ is any which function $\mathbb N\to\mathbb R$.

Then (... bla bla bla, diagonal construction ...) and that produces a real number $x_f$ (which depends on $f$) such that $f(n)\ne x_f$ for all $n$.

In other words $f$ is not surjective (and so not bijective either).

There's not even any proof by contracdiction here, just a direct argument that every function $\mathbb N\to\mathbb R$ is non-surjective.


(Part 2 of your question has been asked and answered many times: the function you're describing fails to hit any real number with a non-terminating decimal expansion, such as $\tfrac13$ or $\sqrt{2}$ or $\pi$).

Troposphere
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