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