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A certain section of the chapter "The Axiom of Choice" on "Naive Set Theory" got me confused:

"[...] The assertion is that a set is infinite if and only if it is equivalent to a proper subset of itself. The "if" we already know; it says merely that a finite set cannot be equivalent to a proper subset. To prove the "only if," suppose that $X$ is infinite, and let $v$ be a one-to-one correspondence from $\omega$ into $X$. If $x$ is in the range of $v$, say $x=v(n)$, write $h(x)=v(n^{+})$; if $x$ is not in the range of $v$, write $h(x)=x$. It is easy to verify that $h$ is a one-to-one correspondence from $X$ into itself. Since the range of $h$ is a proper subset of $X$ (it does not contain $v(0)$), the proof of the corollary is complete. The assertion of the corollary was used by Dedekind as the very definition of infinity."

(The corollary would follow from "every infinite set has a subset equivalent to $\omega$".)

The question is: how can $h$ be a one-to-one correspondence from $X$ into itself if its range does not contain an element (namely, $v(0)$) that is in $X$?

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Note that the word "equivalent" requires context, literally it would just be interpreted as "satisfying some equivalence relation", but what the nature of this relation is not automatically understood.

In the context of cardinality it means to have a bijection. Namely an infinite set has a bijection with a proper subset of itself. Of course this requires the axiom of choice (or rather a small fragment of it) to hold. This characterization is known as Dedekind-infinite. Without the axiom of choice it is consistent that there are sets which are infinite (in the sense that they are not with bijection with any finite ordinal), but they are not Dedekind-infinite.

For example note that $f(n)=n+1$ is a bijection from $\omega$ into $\omega\setminus\{0\}$. It shows that $\omega$ is a Dedekind-infinite set. For the more general case see my answer for Equivalent characterisations of Dedekind-finite proof.

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Asaf Karagila
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  • In this bijection the range of $f$ is $\omega \setminus {0}$, but in the case I mentioned $h$ is supposed to be a one-to-one correspondence from $X$ into itself, and I presumed it should not exclude of its range any elements of $X$. (I'm considering "one-to-one correspondence" to be a bijection) –  Dec 03 '12 at 17:46
  • No, this is another issue with using prosaic language. $f$ is a bijection between $\omega$ and a proper subset of $\omega$. Note that by your interpretation, the book says that all infinite sets are countable. – Asaf Karagila Dec 03 '12 at 17:47
  • Is it wrong to consider "a one-to-one correspondence from $X$ into itself" a bijection from $X$ into $X$? –  Dec 03 '12 at 18:08
  • Yes. It is wrong. Correspondence is an ambiguous term. Does it mean a function? Does it mean a bijection? If it says a bijection then the book claims, by saying "every infinite set $X$ has a one-to-one correspondence from $\omega$ into $X$" that every infinite set is countable. This can't be right... you should understand "one-to-one correspondence" simply as an injection. – Asaf Karagila Dec 03 '12 at 18:10
  • I think I got it. Maybe with "one-to-one correspondence of $X$ into itself" he meant "an injection of $X$ into itself", as you pointed (as opposed to a possible "one-to-one correspondence of $X$ onto itself"). Implicitly, he concludes that indeed there's a "one-to-one correspondence of $X$ onto $X \setminus {v(0)}$, an "equivalence" is this sense, to a proper subset, which is sufficient to prove the initial assertion. –  Dec 03 '12 at 19:47
  • @Fred: Exactly. – Asaf Karagila Dec 03 '12 at 19:48
  • Can someone give an example of a infinite set which is not Dedekind infinite? – Agile_Eagle Feb 19 '21 at 09:11
  • @Agile_Eagle: Example? No. Since ZF is consistent with both "there is an infinite Dedekind-finite set" and with "every infinite set is Dedekind-infinite" (although not with both at the same time, of course) that means that there's no explicit example. – Asaf Karagila Feb 19 '21 at 09:14
  • Thanks for replying (to such a old post) :) I don't know what AOC is, and was kinda confused why Rudin doesn't put iff for this statement – Agile_Eagle Feb 19 '21 at 09:15
  • I guess this is slightly outside the ambit of a math sophomore. I have no idea what ZF is supposed to mean? – Agile_Eagle Feb 19 '21 at 09:16
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    @Agile_Eagle: ZF is the Zermelo–Fraenkel set theory. In a deep sense, ZFC (which is ZF+AC) can be seen as the basis of mathematics, and as a rough intuition, objects we can prove to exist in ZFC, but not in ZF are "non constructive", i.e. they don't have a really explicit description. But the same can be said about objects that ZFC proves don't exist (e.g. infinite Dedekind-finite sets), but ZF does not prove their nonexistence. – Asaf Karagila Feb 19 '21 at 09:24