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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!

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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?

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  • $\begingroup$ The product of sets is empty iff at least on of the sets is empty. $\endgroup$ – William Elliot Sep 29 '17 at 1:40
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    $\begingroup$ @WilliamElliot How do you prove that if a product is empty, then one of the sets is empty? $\endgroup$ – John Griffin Sep 29 '17 at 1:47
  • $\begingroup$ 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) $\endgroup$ – kyp4 Sep 29 '17 at 3:51
  • $\begingroup$ 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) $\endgroup$ – kyp4 Sep 29 '17 at 3:55
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    $\begingroup$ Lemma 1 is perfectly correct except that you need to assume $B$ is nonempty for the reverse direction. $\endgroup$ – Eric Wofsey Sep 29 '17 at 4:39

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