Edit: As pointed out in a comment and in an answer, we're talking about a metric space with distance $d$, $\overline{A}$ is the closure of $A$ and $U$ is a universe.
I'm required to write the proof for:
Prove that if $ A \subset U $ is a set then $\overline{A} $ is closed.
What I came up is this:
Assume $A \subset U $.
WTS: $\overline{A}$ is closed. $\overline{A}$ is closed if its complementary $(\overline{A} ) \, ^ C$ is open.
Take any $x \in (\overline{A} ) \, ^ C $.
Recall $ \overline{A} \equiv \{y\in U | \forall \epsilon > 0, > B_{\epsilon}(y) \cap A\neq \varnothing \}. $
Note that $B_{\epsilon}(y) \cap A \neq \varnothing \iff B_{\epsilon}(y) \not\subset A ^C $.
Then we can write $ \overline{A} \equiv \{y\in U | \forall \epsilon > 0, B_{\epsilon}(y) \not\subset (\overline{A} )^C \}. $
So, $x \in (\overline{A} )^C \iff x \notin \overline{A} \iff \neg \, \{x\in U | \forall \epsilon > 0, B_{\epsilon}(x) \not\subset A^C \} \iff $
$x\in U s.t. \exists \epsilon > 0, B_{\epsilon}(x)\subset A^C. $
Then I'm stuck, because I actually needed to get to $x\in U s.t. \exists \epsilon > 0, B_{\epsilon}(x)\subset (\overline{A})^C. $ I've probably made some mistake or I miss something obvious on how to go on. Can someone help me? Thanks in advance.
I’m assuming $\overline{A}$ is the closure of $A$ and $U$ establishes some sort of universe. I’m also assuming we’re talking about a metric space with metric $d$.
If $y \notin \overline{A}$, then by definition $ \exists \epsilon \gt 0$ with $B(y, \epsilon) \cap A = \emptyset$. Let $d(y, z)= \gamma \lt \epsilon$. Let $\delta=(\epsilon-\gamma)/2$. Then by the triangle inequality, $B(z, \delta) \subseteq B(y, \epsilon)$, so $\forall~z\in B(y, \epsilon), z \notin \overline{A}$ and $B(y, \epsilon) \cap \overline{A} = \emptyset$.
There’s a more general definition of closure that doesn’t rely on the use of a metric and that makes the proof easier. You can define the closure of a set (in a given topology) as the intersection of all closed sets containing that set.