I am trying to proof $\overline{A \cup B}= \overline{A} \cup \overline{B} $, where $\overline{A}$ is the topological closure of A.
I did some thinking and I came up with this.
My Question is: Is my proof correct?
First I am trying to proof $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$
Let $x \in \overline{A \cup B}=(\cap_{A \cup B \subseteq P, P closed}P)$ For every P that statisfies the condition below the cap $x$ is in P. Furthermore $A \subseteq \overline{A}$ and $B \subseteq \overline{B}$, as result we get
$A \cup B \subseteq \overline{A} \cup \overline{B}$.
As union of two closed sets $\overline{A} \cup \overline{B}$ is obviously closed. So the condition for P is statisfied and as consequence $x \in \overline{A} \cup \overline{B}$
The Second direction is:
$\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$
Let $x \in \overline{A} \cup \overline{B}$, without loss of generality let $x \in \overline{A}=(\cap_{A \subseteq P, P closed}P)$
This means x is in every P that statisfies A $\subseteq$ P and P is closed.
$\overline{A \cup B}:=(\cap_{A \cup B \subseteq H, H closed}H)$
trivially $A \subseteq A \cup B$, so if $A \cup B$ is a subset of a set $H$ that is closed, so is $A$, which means that all $H$ satisfy the conditions for $P$. This means $x \in H$ for all $H$ that satisfies $A \cup B \subseteq H$, $H$ closed. So as a result $x \in \overline{A \cup B}$
Best Answer
What you did looks correct.
However, it can be greatly simplified if you know (or prove) that if a subset $A$ is included in a closed set $C$, then so is $\overline{A}$.
As an example, to prove $$\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$$ you could just say:
As $A,B \subseteq \overline{A} \cup \overline{B}$, we have $A \cup B \subseteq \overline{A} \cup \overline{B}$. And as $\overline{A} \cup \overline{B}$ is closed as a (finite) union of closed subsets, we get $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$.