There is a theorem from Rudin's $ \textit{Principle of Mathematical Analysis} $ which says that if $ \displaystyle \{E_{n}\}, \; n = 1, 2, 3, \dots $ is a sequence of countable sets, then their union $ \displaystyle S = \bigcup\limits_{n = 1}^{\infty} \; E_{n} $ is at most countable. I have successfully understood the proof for this theorem. (The notion of a countable set used in Rudin's book is slightly different from the one I find on Wikipedia, please see below for the clarification of the concept.)
Rudin then goes on to this theorem: Let $ A $ be an at most countable set and for every $ \alpha \in A, \; B_{\alpha} $ is at most countable. Prove that $ \displaystyle T = \bigcup\limits_{\alpha \; \in \; A} \; B_{\alpha} $ is at most countable. He says that since $ T $ is equivalent to a subset of $ S $ from above, this theorem must be true, but I can't see why it is the case. Specifically, if all the $ B_{\alpha} $s are finite or if they are all countable, then $ T $ is finite or countable and hence $ T $ is at most countable. But what happens when some $ B_{\alpha} $s are finite and some are countable? In addition, I don't see the reason why the set $ A $ must be at most countable?
Any help would be appreciated!
Some definitions used in Rudin's book: Let $ M $ be a set and $ J_{n} $ be the set of all positive integers up to $ n, $ then $ M $ is finite if there is a bijection from $ M $ to $ J_{n} $ for some positive integer $ n, M $ is countable if there is a bijection from $ M $ to $ \mathbb{Z}^{+}, $ and $ M $ is at most countable if $ M $ is finite or countable.
Best Answer
Terminology note: most authors use "countable" to mean what Rudin calls "at most countable." I use Rudin's terminology below, for clarity, but I just wanted to point this out to forestall any confusion down the road if you look at other sources.
Every finite set is contained in a countably infinite set - just take its union with $\mathbb{N}$.
So suppose I have a collection of sets $B_\alpha$ ($\alpha\in A$), where $A$ is countable; and each $B_\alpha$ is either finite or countable.
Let $C_\alpha=B_\alpha\cup\mathbb{N}$. Then each $C_\alpha$ is countable, so $\bigcup_{\alpha\in A} C_\alpha$ is countable.
But $\bigcup_{\alpha\in A} B_\alpha\subseteq \bigcup_{\alpha\in A} C_\alpha$ since $B_\alpha\subseteq C_\alpha$ for all $\alpha\in A$, and any subset of a countable set is at most countable.
Also, note that even if each $B_\alpha$ is finite, that does not mean that their union is finite! E.g. let $A=\mathbb{N}$ and set $B_\alpha=\{\alpha\}$.
As to why the index set $A$ needs to be at most countable: otherwise the theorem is false!
If $A$ is uncountable, let $B_\alpha=\{\alpha\}$ for each $\alpha\in A$. Then each $B_\alpha$ is at most countable, but $\bigcup_{\alpha\in A}B_\alpha=A$ is uncountable!