I've been reading Apostol text and I have this question.
He states that due to thm 2.25 and 2.26, which is shown below enter image description here

enter image description here

So from these two theorems, this result is immediate If F is a countable collection of countable sets, then the union of all sets in F is also a countable set.

But I don't get how this two theorem implies this. Can someone elaborate more for me?? I know that if $$ F = \{ F_1,F_2, ... \} $$ is a countable collection of countable set, so $$ F_1 = \{ A_{11}, A_{12}, ... \}, F_2 = \{ A_{21}, A_{22}, ... \},... $$

Then by thm 2.26 and thm 2.25, $$ \bigcup A_{1i} $$ is countable and so on for other $ \bigcup A_{ni}$.

But how does this implies the result?? thank you

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    $\begingroup$ The idea is you take an arbitrary collection of sets (the $A_i$) and use 2.26 to turn it into a disjoint collection ($B_i$) which has the same union. Then 2.25 allows you to conclude that this disjoint union is countable. $\endgroup$ – Funktorality Apr 4 '16 at 6:20
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    $\begingroup$ If you're given countably many countable sets $A_i$, then you can find countably many disjoint countable sets $B_i$, such that $\displaystyle \bigcup A_i = \displaystyle \bigcup B_i$, by Theorem 2.26. GIven these countably many disjoint countable sets, $\displaystyle\bigcup B_i$ is countable (by Theorem 2.25). Hence, $\displaystyle\bigcup A_i$ is countable, since it's the same set as $\displaystyle\bigcup B_i$. $\endgroup$ – Christopher Carl Heckman Apr 4 '16 at 6:20

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