Characterization of Closure by Open Sets

Theorem
Let $T = (S,\tau)$ be a topological space.

Let $A$ be a subset of $T$.

Let $x$ be a point of $T$.

Then $x \in \operatorname{Cl} A$ iff for every open subset $U$ of $T$ if $x \in U$, then $A \cap U \neq \emptyset$.

Proof
To prove first implication assume $x \in \operatorname{Cl} A$.

To prove indirectly, assume that there exists an open subset $U$ of $T$ such that $x \in U$ and $A \cap U = \emptyset$.

$A` \cup U` = \left({A \cap U}\right)` = S$.

$A \subseteq U`$
 * Let $y$ be a point.
 * Assume $y \in A$.
 * Then $y \in S$.
 * Then $y \in A`$ or $y \in U`$.
 * Hence $y \in U`$.

$\operatorname{Cl} \left({U`}\right) = U`$.

Then $\operatorname{Cl} A \subseteq U`$.

Hence contradiction.

Assume for every open subset $U$ of $T$ if $x \in U$, then $A \cap U \neq \emptyset$.

To prove indirectly, assume $x \not\in \operatorname{Cl} A$.

Then $x \in \left({\operatorname{Cl} A}\right)`$.

Then $A \cap \left({\operatorname{Cl} A}\right)` \neq \emptyset$ by assumption.

$A \cap A` = \emptyset$. $A \subseteq \operatorname{Cl} A$.

Then $\left({\operatorname{Cl} A}\right)` \subseteq A`$.

Hence contradiction.