Countable collection of open sets.

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.

• Looks fine, but you don't need the contradiction. $N_r(x)\subset \cup_{a} E_a$ by definition of union. – TheManWhoNeverSleeps Jan 12 at 0:45
• Countable is not required. Your proof works with any collection. – herb steinberg Jan 12 at 0:49
• collection means finite or infinite? – mathsssislife Jan 12 at 0:49
• @spaceisdarkgreen may I please have a counter example for my implication and a possible proof for yours, please? – mathsssislife Jan 12 at 0:53
• Let the total space be $\{1,2,3,4,5,6\},$ $A=\{1,2\}$ and $B=\{3,4\}.$ Then $A\cup B=\{1,2,3,4\}$ and $A^c\cap B^c = \{5,6\}.$ The set $\{1,4,5\}$ is not a subset of $A\cup B,$ nor is it contained in $A^c\cap B^c.$ – spaceisdarkgreen Jan 12 at 0:58

1. 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.

2. 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.

3. 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.

• One question, when you say a "family" or "collection", does that include an infinite amount? And many thanks for your reply! – mathsssislife Jan 12 at 1:13
• @mathsssislife Sure, $A$ can be any (finite or infinite) index set. A collection is arbitrary (a family is synonymous, there is no difference). – Henno Brandsma Jan 12 at 1:14

Your proof is not quite clear. You are lost after the first line into irrelevant assumptions and conclusions. Note that you want to show the union is open therefore you pick an arbitrary point in the union which is necessarily in one of the sets. Since this set is open there is an open ball centered at the point and contained in the set which is contained in the union.

Now you can polish your proof using mathematical notations.

• What about my second proof? – mathsssislife Jan 12 at 0:51
• May you please explain why you consider my assumptions to be irrelevant, because if the neighborhood is not contained in the union of the countable number of open sets then that would mean $N_r(p)$ $\subset$ ($\bigcup\limits_{a}E_{a}$) $^c$ which is what I said. – mathsssislife Jan 12 at 0:55
• There is one mistake in your second proof. Make sure that you check your statements carefully. – Mohammad Riazi-Kermani Jan 12 at 1:00
• Well, it is already an irrelevant assumption in the statement of the theorems that the union (resp. intersection) is countable. – Henning Makholm Jan 12 at 1:03
• Now your first proof is correct without the contradiction part. – Mohammad Riazi-Kermani Jan 12 at 1:06