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The question is: Prove if $A_1, . . . , A_m$ are each countable sets then $A_1∪ · · · ∪ A_m$ is countable.

My attempt: First consider two countable sets $A_1$ and $A_2$. And let $B_2=A_2/A_1$, so $A_1$ and $B_2$ is disjoint. Then $A_1 \cup A_2 = A_1 \cup B_2$. Now since $A_1$ and $B_2$ are countable, $A_1=\left\{a_1, a_2, .. \right\}$ and $B_2=\left\{b_1, b_2, .. \right\}$. Then $A_1 \cup B_2=\left\{ a_1, b_1, a_2, b_2, ..\right\}$. Now if we define a function $f:\mathbb{N} \rightarrow A_1 \cup B_2$ in the following way,

$$ f(n)= \left\{\begin{matrix} a_\frac{n+1}{2}& \text{if n is odd} \\ b_n& \text{ if n is even} \\ \end{matrix}\right. $$ Now we can see that $A_1 \cup B_2= A_1 \cup A_2$ is countable. Now using induction we can show that $A_1∪ · · · ∪ A_m $ is countable. Suppose $A_1∪ · · · ∪ A_m$ is countable. Then $(A_1∪ · · · ∪ A_m )\cup A_{m+1}$ is countable since we know that the union of two countable set is countable.

Is that the right way to prove it?

Toneri Otsutsuki
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  • No. that just shows that a finite union of countable sets is countable. – lulu Jun 05 '23 at 11:15
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    The underlying question has been asked many times here...this is one duplicate. this is another. – lulu Jun 05 '23 at 11:17
  • @lulu but the question didn't ask for infinite union of countable sets? – Toneri Otsutsuki Jun 05 '23 at 11:36
  • Of course it did. If you wanted all your sets to be finite you should have said finite instead of countable. – lulu Jun 05 '23 at 11:48
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    I am referring to the question in the header. If you meant to ask something else, please edit for clarity. – lulu Jun 05 '23 at 11:50
  • @lulu I have edited it. – Toneri Otsutsuki Jun 05 '23 at 11:54
  • Ok, well, as I said initially, the argument you gave is fine for that case. Alternatively, you could use the same technique you applied in the case $m=2$. Just put set $A_i$ in the slots $\equiv i \pmod m$ and alter your function accordingly. – lulu Jun 05 '23 at 12:01
  • To simplify it further, you really need to prove that the union of two countable sets is countable. The finite unions, then just follow by induction. – Mr.Gandalf Sauron Jun 05 '23 at 12:04

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