Intersection of Open Set with Closure of Set is Subset of Closure of Intersection

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $H \in \tau$ be an open set of $T$.

Let $K \subseteq S$ be an arbitrary subset of $S$.

Then:
 * $H \cap \map \cl K \subseteq \map \cl {H \cap K}$

where $\cl$ denotes set closure.

Proof
Let $x \in H \cap \map \cl K$.

Then:
 * $x \in H$

and:
 * $x \in \map \cl K$

Let $N_1$ be an arbitrary neighborbood of $x$.

Because $x \in H$, there exists a neighborbood $N_2$ of $x$ entirely within $U$.

Let $N_3 = N_1 \cap N_2$.

By Intersection of Neighborhoods in Topological Space is Neighborhood, $N_3$ is also a neighborbood of $x$.

Because $N_1 \cap N_2 \subseteq N_2$, $N_3$ is also contained entirely within $U$.

Because $x \in \map \cl K$, by definition of closure, there exists $y \in N_3$ which is also in $K$.

From $x \in N_3$ we know that $y \in H$.

Thus we have demonstrated the existence of $y \in N_1$ that is in $H \cap K$.

As $N_1$ is arbitrary, it follows that for every neighborbood of $x$ there exists $y \in N_1$ such that $y \in H$ and $y \in K$.

That is:
 * $x \in \map \cl {H \cap K}$

Hence the result, by definition of subset.