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Let $X$ be a topological space, $I$ be an infinite set and $(S_i)_{i \in I} \subset X$ be a family of subsets of $X$. I want to show:

\begin{equation} \overline{\bigcup_{i \in I} S_i} = \overline{\bigcup_{i \in I} \overline{S_i}} \end{equation}

I have the feeling, that this proof should be really easy, but somehow I don't get it.

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  • $\begingroup$ Hint: to show $\overline A\subseteq B$, it's enough to show $A\subseteq B$ and $B$ is closed. $\endgroup$
    – Wojowu
    May 31, 2018 at 10:38

2 Answers 2

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For all $i$, $S_i \subseteq \overline{S_i}$, so

$$\bigcup_i S_i \subseteq \bigcup_i \overline{S_i}$$

and this inclusion is preserved by taking closures on both sides, hence

$$\overline{\bigcup_i S_i} \subseteq \overline{\bigcup_i \overline{S_i}}$$

On the other hand, for each (fixed) $i$:

$$S_i \subseteq \bigcup_i S_i$$ and again talking closures on both sides, we get:

$$\overline{S_i} \subseteq \overline{\bigcup_i S_i}$$ and as $i$ was arbitary, and the right hand side is a fixed set:

$$\bigcup_i \overline{S_i} \subseteq \overline{\bigcup_i {S_i}}$$ now, the right hand side is already closed so, taking closures on both sides again:

$$\overline{\bigcup_i \overline{S_i}} \subseteq \overline{\overline{\bigcup{S_i}}} = \overline{\bigcup{S_i}}$$

and we have shown both inclusions. We don't need the size of $I$; it could well be finite too.

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Left side subset right side is easy.
The key for the reverse inclusion is for all i, $\overline{S_i} \subseteq \overline{\cup_i S_i}.$

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