Closure of Intersection is Subset of Intersection of Closures

Theorem
Let $T$ be a topological space.

Let $I$ be an indexing set.

Let $\forall i \in I: H_i \subseteq T$.

Then $\operatorname{cl}\left({\bigcap_I H_i}\right) \subseteq \bigcap_I \operatorname{cl}\left({H_i}\right)$.

Proof
Since $\bigcap_I \operatorname{cl}\left({H_i}\right)$ is an intersection of closed sets, it is closed, from Topology Defined by Closed Sets.

Also, it contains $\bigcap_I H_i$ and so by the main definition of closure also contains $\operatorname{cl}\left({\bigcap_I H_i}\right)$.

Note
Equality does not generally hold.

Take for example:
 * $H \subseteq \R: H = \left({0 \, . \, . \, 2}\right) \cup \left({3 \, . \, . \, 4}\right)$;
 * $K \subseteq \R: K = \left({1 \, . \, . \, 3}\right)$

where $\R$ is under the usual topology.


 * $H \cap K = \left({1 \, . \, . \, 2}\right)$;
 * $\operatorname{cl}\left({H \cap K}\right) = \left[{1 \, . \, . \, 2}\right]$;
 * $\operatorname{cl}\left({H}\right) = \left[{0 \, . \, . \, 2}\right] \cup \left[{3 \, . \, . \, 4}\right]$;
 * $\operatorname{cl}\left({K}\right) = \left[{1 \, . \, . \, 3}\right]$;
 * $\operatorname{cl}\left({H}\right) \cap \operatorname{cl}\left({K}\right) = \left[{1 \, . \, . \, 2}\right] \cup \left\{{3}\right\}$.

Thus $\operatorname{cl}\left({H \cap K}\right) \ne \operatorname{cl}\left({H}\right) \cap \operatorname{cl}\left({K}\right)$.