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According to the Set Theory book I am studying, the Axiom of Choice is required to prove that the union of a countable number of at most countable sets is itself at most countable. I seem to have proven this without using choice but I am certain that I made a mistake somewhere! As I use a lot of custom macros, I posted the proof here. I hope this isn't breaking any rules.

I'm pretty sure the error is in Lemma 1 and something that I am doing assumes/requires choice (or well-ordering) but I do not know what. Any help in pointing out the problem would be much appreciated!

John Griffin
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kyp4
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1 Answers1

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"First, since each $A_n$ is at most countable, we have $|A_n| \le \aleph_0$ so that by Lemma 1 there is a function $f_n$ from $\mathbb{N}$ onto $A_n$ since clearly $\mathbb{N}$ is well-ordered."

You are using countable choice to choose these $f_n$'s. All you know is that the set $$ F_n:= \{f:\mathbb{N}\to A_n \mid f\ \text{is onto}\} $$ is nonempty for each $n$. How do you select a specific $f_n$ from each of them? In other words, how do you know that $$ \prod_{n\in\mathbb{N}} F_n $$ is nonempty?

John Griffin
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  • The product of sets is empty iff at least on of the sets is empty. – William Elliot Sep 29 '17 at 01:40
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    @WilliamElliot How do you prove that if a product is empty, then one of the sets is empty? – John Griffin Sep 29 '17 at 01:47
  • Does this mean that Lemma~1 is actually correct? I am a little confused by your answer due to not having studied the AoC in depth yet (as it's the next chapter in the book). I did just read/skim that chapter and it confirmed a lot of things I suspected about it. In that chapter they prove this theorem using choice and do mention choosing the sequences as you do. (1/3) – kyp4 Sep 29 '17 at 03:51
  • What I do not understand is why do we need the AoC to choose a particular sequence $f_n$ but we evidently do not need to choose among the possible sequences that enumerate the countable system of sets (I write it as a sequence in my proof but the book leaves it as a general countable system). It seems to be sufficient here to say that such a sequence exists even though there could be many such sequences. (2/3) – kyp4 Sep 29 '17 at 03:55
  • In general it seems that sometimes in proofs we can simply say that something exists (even when there may be more than one of those things) and be fine and other times we require the use of the AoC. It's not clear to me what the difference is between these two situations. If anyone knows of any discussions of this I'd love to read it, the book I'm using doesn't seem to discuss this distinction at all. (3/3) – kyp4 Sep 29 '17 at 03:56
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    Lemma 1 is perfectly correct except that you need to assume $B$ is nonempty for the reverse direction. – Eric Wofsey Sep 29 '17 at 04:39
  • @kyp4 Lemma 1 is fine if you consider Eric's correction. You can actually prove the reverse direction without using the fact that $A$ is well-ordered, because you can choose $a$ in the range of $f$. This gets to your main point of confusion. You can select an element from a nonempty set without the axiom of choice. In fact you can select an element from each set in a finite collection of nonempty sets. The axiom of choice comes into play whenever one wishes to select an element from each set in an infinite family of nonempty sets. This is precisely want we need in order to choose the $f_n$'s. – John Griffin Sep 29 '17 at 05:06
  • @JohnGriffin. Ok, it is equivalent to AxC. – William Elliot Sep 29 '17 at 06:17
  • @kyp4 Also see Asaf's answer here: https://math.stackexchange.com/questions/85153/how-do-i-choose-an-element-from-a-non-empty-set – John Griffin Sep 29 '17 at 21:00
  • I've done some reading, including in a number of related questions here, and I think I understand the difference between when the AoC is needed and when it's not, namely you need it if you're making an infinite number of choices. One question I do have is that, in my proof, I construct the specific onto function in Lemma 1 so that a choice is not necessary if we use that specific function. However, then the choice is just pushed back because we must make a choice of which injective function to use from which to construct the onto function. Hence the AoC is still required! Is this correct? – kyp4 Oct 01 '17 at 15:03
  • @kyp4 Yes it's correct! – John Griffin Oct 01 '17 at 15:07
  • Okay, one final question I was just thinking about. Even if the sets in your system were well-orderable, wouldn't you still have to choose an infinite number of well-orderings (one for each set), hence the AoC is still needed? On the other hand, if the sets have a given well-ordering (i.e. they are well-ordered rather than merely well-orderable) then you don't need the AoC because the well-ordering serves as a choice function. Is this correct? – kyp4 Oct 01 '17 at 15:57
  • @kyp4 Yes that's also correct! One possible way around it is if you know they are simultaneously well-orderable; i.e., their union can be well-ordered. – John Griffin Oct 01 '17 at 16:55
  • Right because then you can choose a single well-ordering and use that to make a choice function (select the least element in each set according to the well-ordering since each set is a subset of the well-ordered union). Interesting! – kyp4 Oct 01 '17 at 19:15