Now here's another question during elemental set theory. I met the subject of indexed sets, and of course the distributive law holds as follows:
$\displaystyle A\cap\left(\bigcup_{i\in I}A_i\right)=\bigcup_{i\in I}(A\cap A_i)$ and $\displaystyle A\cup\left(\bigcap_{i\in I}A_i\right)=\bigcap_{i\in I}(A\cup A_i)$
So I came up with a proof of the first one:
\begin{alignat*}{2} x\in A\cap\left(\bigcup_{i\in I}A_i\right)&\Leftrightarrow (x\in A)\land(\exists i\in I:x\in A_i) \\ &\Leftrightarrow\exists i\in I:(x\in A)\land (x\in A_i) \\ &\Leftrightarrow\exists i\in I:x\in A\cap A_i &&\Leftrightarrow x\in\bigcup_{i\in I}(A\cap A_i) \end{alignat*}
But I think the second $\Leftrightarrow$(call it Eq. 1) is a bit weak. Then some reading told me that the $\exists$ symbol is actually the same as multiple $\lor$s.
Now if the index set $I$ was finite, I could draw a truth table(with some effort, of course) and find that Eq. 1 is true. If $I$ was countable, then I could possibly proceed with induction. But there is no restriction of $I$, so I'm not sure how I'm supposed to handle this. What should I do? Thanks in advance.
The cardinality of $I$ is irrelevant. Either some $i\in I$ satisfies $x\in A_i$, in which case that choice of $i$ makes both sides of the second $\iff$ true, or none does, in which case both sides of the second $\iff$ are false.