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:
 * $\displaystyle \left({\bigcap_I H_i}\right)^- \subseteq \bigcap_I H_i^-$

where $H_i^-$ denotes the closure of $H_i$.

Proof
Since $\displaystyle \bigcap_I H_i^-$ 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 $\displaystyle \left({\bigcap_I H_i}\right)^-$.

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.


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

Thus $\left({H \cap K}\right)^- \ne H^- \cap K^-$.