Prove that $\overline{A\cup B} = \overline{A}\cup\overline{B}$ and $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$
My attempt:
$x\in\overline{A\cup B}$ iff for every open set $U$ containing $x$, $U\cap\ (A\cup B)\neq\varnothing$. This happens iff $(U\cap A)\cup (U\cap B)\neq\varnothing$, and this happens iff $U\cap A\neq\varnothing$ or $U\cap B\neq\varnothing$, and this happens iff $x\in\overline{A}$ or $x\in\overline{B}$, and this is true iff $x\in\overline{A}\cup\overline{B}$. So $\overline{A\cup B} = \overline{A}\cup\overline{B}$
If $x\in A\cap B$, then $x\in A$ and $x\in B$, which implies $x\in\overline{A}$ and $x\in\overline{B}$ (since closure of $E$ is the smallest closed set containing $E$), so $x\in\overline{A}\cap\overline{B}$. This shows that $A\cap B\subseteq\overline{A}\cap\overline{B}$, and since $\overline{A}$ and $\overline{B}$ are both closed, $\overline{A}\cap\overline{B}$ is closed. But by definition, $\overline{A\cap B}$ is a subset of every closed set containing $A\cap B$, so $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$.
Do my proofs look correct?
Best Answer
The reasoning is not quite correct. The statement
can be expressed as
While
is
The second statement clearly implies the first one, so $$x\in\overline A\vee x\in\overline B\implies x\in\overline{A\cup B}$$ In order to show the other direction, I recommend that you try and prove the contrapositive. So you start by assuming that
and then seek to obtain from this the negation of the first statement. Here you will actually need some topology.
Ironically you used topology in the second proof where it can most efficiently be done the way you proved the $\overline{A}\cup\overline B\subseteq\overline{A\cup B}$ direction.