# Closure of Intersection is Subset of Intersection of Closures

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## Theorem

Let $T$ be a topological space.

Let $I$ be an indexing set.

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

Then:

- $\ds \map \cl {\bigcap_I H_i} \subseteq \bigcap_I \map \cl {H_i}$

where $\map \cl {H_i}$ denotes the closure of $H_i$.

## Proof

Since $\ds \bigcap_I \map \cl {H_i}$ is an intersection of closed sets, it is closed, from Topology Defined by Closed Sets.

Also, it contains $\ds \bigcap_I H_i$ and so by the main definition of closure also contains $\ds \map \cl {\bigcap_I H_i}$.

$\blacksquare$

## Also see

- Closure of Intersection may not equal Intersection of Closures, which shows that equality does not generally hold.

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.17$