I am aware of the proofs of the following two theorems; however, I would like to know if my proof, in terms of logic is correct (and in general as well).
Let $E_a$ be a countable collection of open sets then their union is an open set.
Proof: Since $\forall a$, $E_a$ is open, it follows that $x\in$ $E_a$ $\implies$ $\exists r>0 : N_r(x)\subset E_a$. Hence $N_r(x)\subset\bigcup\limits_{a}E_a$, because $N_r(x)\nsubseteq\bigcup\limits_{a}E_a$ $\implies$ $N_r(x)\subset\bigcap\limits_{a}E_a^c$, hence $\forall a : N_r(x)\subset E_a^c$ which is a contradiction.
Let Let $E_a$ be a countable collection of closed sets then their intersection is a closed set.
proof: Since $\forall$ a $C_a$ is closed $\implies$ $\forall$ a $C_a^c$ is open. Let x be a limit point of $\bigcap\limits_{a}C_{a}$ then since x$\in$ $C_a^c$ $\forall$ a it follows that x$\in$ $\bigcap\limits_{a}C_{a}$, hence it is closed.
Are my proofs correct? Any feedback is welcome, please.
Best Answer
The restriction to countable unions of open sets or countable intersections of closed sets is unnecessary. Nowhere do you use that countability, as you could have noticed.
The proof for open sets is easier than you make it out to be: let $E_a, a \in A$ be a family of open sets and $E:= \bigcup_{a \in A} E_a$ be their union. We want to show that every point of $E$ is an interior point, so let $x \in E$. By the definition of a union we know there is some $a(x) \in A$ such that $x \in E_{a(x)}$. As the latter set is open there is some $r>0$ such that $N_r(x) \subseteq E_{a(x)}$. Trivially, $E_{a(x)} \subseteq E$ so $N_r(x) \subseteq E$ as well and we are done. $E$ is open.
As to closed sets: We can either use that a set is open iff its complement is closed and then the statement about intersections of closed sets is an immediate consequence of the previous fact from de Morgan’s law:
$$\left(\bigcap_{a \in A} E_a\right)^c = \bigcup_{a \in A} (E_a)^c$$
But as you seem to prefer the limit point definition in your proof: suppose that $E_a, a \in A$ now is a family of closed sets and $E=\bigcap_{a \in A} E_a$ their intersection. If $x$ is a limit point of $E$, then it for any $a$ it is a limit point of $E_a$ (as $E \subseteq E_a$) and so for any $a$ we have $x \in E_a$ as all $E_a$ are closed. Hence $x \in E$ and we are done. This largely corresponds to the idea in your proof, I think, but is more direct.