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:
 * $\displaystyle \operatorname{cl}\left({\bigcap_I H_i}\right) \subseteq \bigcap_I \operatorname{cl}\left({H_i}\right)$

where $\operatorname{cl}\left({H_i}\right)$ denotes the closure of $H_i$.

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

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

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