# Closure in a topological space

Let $$(X, \tau)$$ be a topological space. Let $$A,B$$ be subsets of $$X$$. Show that $$cl(A \cup B)$$ $$\subset$$ $$cl(A)$$ $$\cup$$ $$cl(B)$$

Proof: Let x be in the closure of $$A \cup B$$. That means for every open set U containing x, $$U \cap (A \cup B)$$ is non empty. That means for every open set U containing x either $$U \cap A$$ is non empty or for every open set U containing x $$U \cap B$$ is non empty. Which means that x$$\in cl(A) \cup cl(B)$$

Is this proof correct?

Not really convincing to be honest. Your proof fully depends on $$U$$. In order to prove that $$x\in cl(A)$$ you need to show that for every open set $$U$$ which contains $$x$$ we have that $$U\cap A$$ is not empty. Same thing about $$cl(B)$$. What you showed is that for every open neighborhood $$U$$ of $$x$$ one of the sets $$U\cap A, U\cap B$$ is not empty, but you don't know which one of them. So there might be an open neighborhood $$U_1$$ for which $$U_1\cap A$$ is not empty and another open neighborhood $$U_2$$ for which $$U_2\cap B$$ is not empty. So how exactly do you conclude that $$x$$ is in the closure of $$A$$ or $$B$$?

I would do it like this: suppose $$x\notin cl(A)\cup cl(B)$$. Then there are open neighborhoods $$U,V$$ of $$x$$ such that $$U\cap A,V\cap B$$ are both empty. But then $$U\cap V$$ is an open neighborhood of $$x$$ and its intersection with $$A\cup B$$ is empty. This is a contradiction to $$x\in cl(A\cup B)$$.

I think not, suppose there existed $$U,V$$ open sets containing x so that $$U \cap A=\emptyset$$(but $$U \cap B\neq\emptyset$$) and $$V \cap B=\emptyset$$(but $$V \cap A\neq\emptyset$$).

Then $$x\not\in cl(A)$$ and $$x\not\in cl(B)$$.

Now of course, what you claimed is true, hopefully you can modify your argument to address this.

Edit: However, your proof in Interior topology was correct, then noticying that $$cl(X-A)=X-int(A)$$ you can build a proof for what you want

Not correct (see the other answers).

As a sort of continuation of my answer on your former question now sortlike characterizations of $$\mathsf{cl}(C)$$:

The closure of set $$C$$ is the intersection of all closed sets that contain $$C$$ as a subset.

or:

The closure of set $$C$$ is the smallest closed set that contains $$C$$ as a subset.

Now observe that $$\mathsf{cl}(A)\cup\mathsf{cl}(B)$$ is evidently a closed set that contains $$A\cup B$$ as a subset, so we may conclude immediately that: $$\mathsf{cl}(A\cup B)\subseteq\mathsf{cl}(A)\cup\mathsf{cl}(B)$$

No, it is not correct. You know that every open set $$U$$ to which $$x$$ belongs intersects $$A\cup B$$. It is not obvious just from this that $$U$$ intersects $$A$$ and $$B$$.

Note that if $$x\notin\overline B$$, then, since $$\overline B$$ is a closed set, then $$\overline B^\complement$$ is an open set to which $$x$$ belongs. Therefore, $$U\cap\overline B^\complement$$ intersects $$A\cup B$$, for every open set $$U$$ to which $$x$$ belongs. But it does not intersect $$B$$. So, it intersects $$A$$. So, $$x\in\overline A$$.

As noted bellow it is not correct. It can be that some open sets intersect A but not B and other intersect B but not A. You can build a similar arguement to the one you made, but based on $$x \notin cl(A) \cup cl(B) \Rightarrow x\notin cl(A \cup B)$$. It has been shown here: Proof of $cl(A \cup B)=cl(A) \cup cl(B)$ in the second answer (not the one that was accepted).

In fact it is: 1. $$cl(A \cup B) = cl(A) \cup cl(B)$$

1. $$int(A \cap B) = int(A) \cap int(B)$$

2. $$int(A) \cup int(B) \subseteq int(A \cup B)$$

3. $$cl(A) \cap cl (B) \subseteq cl(A \cap B)$$

You can see the proofs here: https://www.youtube.com/watch?v=T1GXwknimMg&list=PL-_cKNuVAYAX_LKTQPzvg5lKaOI8_EpjL&index=3

starting at 25:30.

The correct conclusion from $$x\in\operatorname{cl}(A\cup B)$$ is

for every neighborhood $$U$$ of $$x$$, then $$U\cap(A\cup B)\ne\emptyset$$

that translates into

for every neighborhood $$U$$ of $$x$$, then either $$U\cap A\ne\emptyset$$ or $$U\cap B\ne\emptyset$$.

Symbolically, denoting by $$\mathscr{U}_x$$ the family of neighborhoods of $$x$$:

(1) $$\forall U \bigl( U\in\mathscr{U}_x \to (U\cap A\ne\emptyset)\lor(U\cap B\ne\emptyset) \bigr)$$

You are stating, instead,

(2) $$\bigl(\forall U (U\in\mathscr{U}_x \to U\cap A\ne\emptyset\bigr) \lor \bigl(\forall U (U\in\mathscr{U}_x \to U\cap B\ne\emptyset\bigr)$$

Can you see the difference between (1) and (2)?

Correct argument. Suppose $$x\notin\operatorname{cl}(B)$$. Then there exists $$U_0\in\mathscr{U}_x$$ such that $$U_0\cap B=\emptyset$$. If $$U\in\mathscr{U}_x$$, then $$(U\cap U_0)\cap B=\emptyset$$; on the other hand, $$U\cap U_0\in\mathscr{U}_x$$ implies $$(U\cap U_0)\cap(A\cup B)\ne\emptyset$$. Therefore $$\emptyset\ne(U\cap U_0)\cap A\subseteq U\cap A$$. This proves $$x\in\operatorname{cl}(A)$$.

Note that (2) is actually true in this particular context, because of the property that $$\mathscr{U}_x$$ is closed under intersections.