# Identity of closure and open sets in a topological space

Using the fact that if $U$ is an open subset of $X$, and $A \subset X$ any subset then:

$\overline{U \cap A} = \overline{U \cap \overline{A}}$

How from this follows that if $F \subset X$ is closed, then:

$int(F \cup int(A))=int(F \cup A)$ ?

Clearly the LHS is always contained in the RHS but I don't see the other inclusion. Any idea?

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$F$ is closed iff its complement is open. So what can you tell about the connection between the interior of one set and the closure of its complement? – Prometheus Aug 8 '11 at 14:52
Are you asking for the proof of the first identity or just about the transition from the first identity to second one? You might have a look here: proofwiki.org/wiki/… – Martin Sleziak Aug 8 '11 at 15:17

Lemma. Let $X$ be a topological space, and let $A\subseteq X$. Then $$\mathrm{int}(A) = X-\overline{X-A}.$$

Proof. Remember that $\mathrm{int}(A)$ is the largest open subset $U$ of $X$ such that $U\subseteq A$, and that $\overline{B}$ is the smallest closed subset $C$ of $X$ such that $B\subseteq C$.

If $C$ is a closed set such that $X-A\subseteq C$, then $X-A\subseteq \overline{X-A}\subseteq C$, hence taking complements we have $X-C\subseteq X-\overline{X-A}\subseteq X-(X-A) = A$. Since $C$ is closed, $X-C$ is open, so $X-C$ is an open set contained in $A$; hence $X-C\subseteq \mathrm{int}(A)$. In particular, $X-\overline{X-A}\subseteq \mathrm{int}(A)$.

Now let $U$ be an open set contained in $A$; then $X-U$ is a closed set that contains $X-A$; since the closure is the smallest closed subset that contains the set, then $\overline{X-A}\subseteq X-U$. So $U=X-(X-U) \subseteq X-\overline{X-A}$. In particular, $\mathrm{int}(A)\subseteq X-\overline{X-A}$. This proves the desired equality. QED

Corollary. $X-\mathrm{int}(A) = \overline{X-A}$.

So we have that \begin{align*} \mathrm{int}(F\cup \mathrm{int}(A)) &= X - \overline{X-(F\cup\mathrm{int}(A))}\\ &= X - \overline{(X-F) \cap (X-\mathrm{int}(A))}\\ &= X - \overline{(X-F) \cap \overline{X-A}}\\ &= X - \overline{(X-F)\cap (X-A)}&\qquad&\text{(using your fact)}\\ &= X - \overline{(X-(F\cup A))}\\ &= \mathrm{int}(F\cup A). \end{align*} as desired.

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