This follows Arturo's answer of this question.

Let $I$ an infinite set and $\{E_i\}_{i\in I}$ a family of finite sets.

It is said that there exists an injection

$$\bigcup_{i\in I} \ E_i \to I.$$

In the comments, an argument with "$\aleph$ numbers" (which I haven't studied yet) is given.

I wondered how one could build this injection explicitely ? I have been trying for some time now, but it is quite hard to have intuition about that.

  • 1
    $\begingroup$ There is no canonical choice for this injection. But there exist many. $\endgroup$ – Rasmus Oct 8 '11 at 20:57
  • $\begingroup$ There need not be any injection if you don't assume the Axiom of Choice. $\endgroup$ – Arturo Magidin Oct 8 '11 at 21:06
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    $\begingroup$ And, generally, when it is said that the existence of something depends on the Axiom of Choice, you shouldn't expect to be able to construct an example explicitly. (It's just barely possible that an explicitly constructible example exists, but it will usually not be provable that said explicit construction is in fact an example. (Exceptions include pathologic examples such as "there exists a natural number $n$ such that $2+n=4$ and the Axiom of Choice holds")). $\endgroup$ – Henning Makholm Oct 8 '11 at 22:26

If the axiom of choice is assumed then we can simply well order everything and just define by induction an injective map.

Simply choose $f_i\colon E_i\hookrightarrow\mathbb N$ which is an injection. Since we assume the axiom of choice we also have:

$$\tag{1}|I|=|I\times I|\ge|I\times\mathbb N|=|I|$$ We can now inject $E_i$ into $\{i\}\times\mathbb N$, and use $(1)$ to obtain a bijection into $|I|$, and thus into $I$.

Without the axiom of choice it is consistent to have a countable union of disjoint pairs, and the result is not at all countable, so there is no such injection.

(This example is of course Russell's well known saying that you need the axiom of choice to choose from infinitely many pairs of socks.)

  • $\begingroup$ All true, but not too kind to throw an $\aleph_0$ at the OP without even an apology, given that he said he was unfamiliar with "$\aleph$ numbers" (but actually the only $\aleph$ in the comment thread he references was $\aleph_0$ itself). @Klaus, $\aleph_0$ is just notation for the cardinal number "countable infinity". It's part of a more involved system of cardinal notations, but that shouldn't be needed here. $\endgroup$ – Henning Makholm Oct 8 '11 at 22:20
  • $\begingroup$ @HenningMakholm Thank you for the precision. I assumed that $(1)$ meant that $I\times\mathbb{N}$ was equipotent to $I$. I found a proof of that fact in Lang's Algebra, Appendix 2. Indeed more involved than I thought it would be (not explicit, axiom of choice). Interesting! $\endgroup$ – Klaus Oct 8 '11 at 22:44
  • $\begingroup$ @Henning: You are correct. I have changed that. $\endgroup$ – Asaf Karagila Oct 8 '11 at 23:06
  • $\begingroup$ @Klaus: I hope that my notation is now somewhat clearer. The fact that $|I|=|I\times I|$ is equivalent to the axiom of choice, and I have given proof somewhere on this site for this. Merely for finite sets I think you can get away with somewhat less than the full axiom of choice, but regardless to that - choice is still necessary, as the second paragraph shows. $\endgroup$ – Asaf Karagila Oct 8 '11 at 23:07
  • $\begingroup$ Well, I knew what $\aleph_0$ is long before I knew what aleph numbers are. For several years I was under the impression that $\aleph_1$ was defined to be $2^{\aleph_0}$, the cardinality of the continuum. In fact the definition of $\aleph_1$ as the cardinality of the set of all countable ordinals goes all the way back to Cantor. $\endgroup$ – Michael Hardy Oct 9 '11 at 0:58

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