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Update

What I was trying to do here was to come up with an axiom that could replace the axiom of countable choice. Although the effort went down in flames, I now more strongly appreciate and respect the set theoretic axiomatic fortress that has been erected over the past century or so.

I tried again (being more careful this time), defining $\text{sub-}\omega$ injections into $\Bbb N$ (as Asaf remarked, bijections can't work) and attempting to write up a new axiom. But all I wound up with is just another way of saying that the union of countable set is countable is because because.

Axiom

Let $(A_k)_{k\in \Bbb N}$ be a family of disjoint sets $A_k$ and $B$ a set such that

$\tag 1 \text{There exists an injective mapping from } A_0 \text{ into } B$ $\tag 2 \text{Every injective map on } \bigcup_{k \lt n} A_k \text{ to } B \text{ can be extended to an injection on } \bigcup_{k \le n} A_k $

Then there exist and injection of $\bigcup_{k = 0}^{\infty} A_k$ into $B$.

Question: Can we use this axiom to prove that the union of countably many countable sets is countable?

My work

I've already sketched out such a proof and will provide my answer here if this question can be answered in the affirmative.

If someone's answer would make my argument redundant, I will accept their answer (besides giving it an upvote).

CopyPasteIt
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1 Answers1

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No it is not quite possible. At least in the naive proof that I guess you're trying to go through. So any proof will require a different level of sophistication.

I am guessing that you are trying to extended injections from the finite unions into $\omega$. But what if at some point your injection is in fact a bijection? Especially when having countably many countable sets, that is more than just a possibility.

So that means that $B$ must be strictly larger than the finite unions. So the one thing you can do is require that $B=\omega_1$, in which case you can always extend the injections. So that means that a countable union of countable sets can only be countable or of size $\aleph_1$.

Alas, in the Truss' model of collapsing a singular cardinals a la Solovay's model, the countable union of countable sets is well-orderable, but it could be of size $\aleph_1$.

To circumvent this, you'll need to also show that there is other family of sets which show that this axiom you're trying to add is false. This is far from obvious, and it will require a deeper understanding of the Truss model which is quite technical.

So what you can prove is that a countable union of well-orderable sets is well-orderable, but it is not clear if you can actually prove the result you're trying to prove.

Asaf Karagila
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  • As I went back to bed I kept thinking about it and just got up to delete this question. As a moderator please delete this question - I hope to come back later to this with, hopefully, more sophistication. – CopyPasteIt Jul 13 '21 at 12:33
  • I copied your answer for notes... – CopyPasteIt Jul 13 '21 at 12:34
  • No. (1) If you delete it, then you can later undelete it and edit it. (2) I spent time writing this answer, and I think it can be helpful for future people who may be perusing this site. You can delete it if you want, and I won't object, but it may or may not affect my decisions to answer your future questions. Take that into account. – Asaf Karagila Jul 13 '21 at 12:35
  • OK - my question looks silly now but that is how we learn...Agree - leaving it up ...upvoting your answer – CopyPasteIt Jul 13 '21 at 12:39
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    Many questions look silly in retrospect. Don't worry about it. That's not a trivial point, as I said in my answer, so it's easy to overlook. – Asaf Karagila Jul 13 '21 at 12:41
  • I accepted your answer and updated my question. – CopyPasteIt Jul 13 '21 at 14:59
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    To your update, yes, that is often the case when you try to prove these sort of things. If you look for a kind of "minimal hypothesis" you end up with "it is because it is" situation. Sometimes there is something smart to say, but more often than not, there's not any. – Asaf Karagila Jul 13 '21 at 15:37