# union of countable sets, Apostol

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

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

• 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. – Funktorality Apr 4 '16 at 6:20
• 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 $\displaystyle\bigcup B_i$. – Christopher Carl Heckman Apr 4 '16 at 6:20