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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} $.

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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.

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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.

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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!)

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