# Union of closures

Let $$X$$ be a topological space $$\mathscr{ B}$$ be a collection of subsets of $$X$$. Show that $$\overline{ \bigcup \limits_{\alpha \in \mathscr{B}} B_\alpha} \subset \bigcup \limits_{\alpha \in \mathscr{B}} \overline{B_\alpha}$$.

My attempt:

If {$${B_\alpha: \alpha \in \mathscr{B}}$$} is a collection of sets in $$X$$. Then $$x \in \overline{ \bigcup \limits_{\alpha \in \mathscr{B}} A_\alpha}$$ then every neighborhood of $$U$$ of $$x$$ intersects $$\bigcup \limits_{\alpha \in \mathscr{B}}$$ Thus, $$U$$ must intersects some $$B_\alpha$$, so $$x$$ must belong to the closure $$\overline{B_\alpha}$$ of some $$B_\alpha$$. Therefore, $$x \in \bigcup \limits_{\alpha \in \mathscr{B}} \overline{B_\alpha}$$.

Your statement is false. Let $$Y$$ be any non-closed subset of $$X$$ and assume that $$X$$ is $$T_1$$. Then each singleton is closed and $$Y=\bigcup_{y\in Y}\{y\}=\bigcup_{y\in Y}\overline{\{y\}}$$. However, since $$Y$$ is not closed, then $$\overline Y\not\subset Y$$. In other words, $$\overline{\bigcup_{y\in Y}\{y\}}\not\subset\bigcup_{y\in Y}\overline{\{y\}}$$.

I suggest that you try to see where is the error in your proof.

In my opinion $$\mathscr{B}$$ has to be finite. Without it, the statement is false. Let me give a counterexample: $$B_k:=[-1,-\frac{1}{k}]\subset\mathbb{R}$$ Then $$\cup_k \overline{B_k}=\cup_k B_k = [-1,0[$$, since the $$B_k$$ are closed. Hence $$0$$ is not in this union. But $$0\in\overline{\cup_k B_k}=[-1,0]$$

If $$\mathscr{B}$$ is finite, then it is correct, since the union of finitely many closed sets is again closed.

I think there is some mistake in your question.

for example, Take $$X=\Bbb{R}$$ and $$\{r_n:n \in \Bbb{N}\}$$ is an enumeration of $$\Bbb{Q}$$. Now take, $$\forall n, B_n:=\{r_n\}$$. Clearly, $$cl(B_n)=\{r_n\}$$ and $$\cup B_n=\Bbb{Q}$$, whose closure is $$\Bbb{R}$$. But $$\cup \overline{B_n}=\Bbb{Q}$$

So, $$\overline{\cup B_n}=\Bbb{R}$$ whereas $$\cup\overline{B_n}=\Bbb{Q}$$

Edit. But the reverse inclusion is true.(Prove it!)