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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    $\begingroup$ Presumably "equivalent to" here means "bijective with". $\endgroup$ – Benedict Eastaugh Dec 3 '12 at 17:13
  • $\begingroup$ The assumption is that $\nu$ is a 1-1 function from $\omega$ into $X$. So in particular, $\nu(0)$ is an element of $X$. Why do you think that $\nu(0)$ is not in $X$? $\endgroup$ – g.castro Dec 3 '12 at 17:25
  • $\begingroup$ @g.castro You're right, I've just edited my question now. $\endgroup$ – user36546 Dec 3 '12 at 17:31

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.

  • $\begingroup$ 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) $\endgroup$ – user36546 Dec 3 '12 at 17:46
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Dec 3 '12 at 17:47
  • $\begingroup$ Is it wrong to consider "a one-to-one correspondence from $X$ into itself" a bijection from $X$ into $X$? $\endgroup$ – user36546 Dec 3 '12 at 18:08
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Dec 3 '12 at 18:10
  • $\begingroup$ 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. $\endgroup$ – user36546 Dec 3 '12 at 19:47

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