Closure of Intersection may not equal Intersection of Closures/Outline

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H_1$ and $H_2$ be subsets of $S$.

Let ${H_1}^-$ and ${H_2}^-$ denote the closures of $H_1$ and $H_2$ respectively.

Then it is not necessarily the case that:

$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$

Proof

From Closure of Intersection is Subset of Intersection of Closures, it is seen that it is always the case that:

$\paren {H_1 \cap H_2}^- \subseteq {H_1}^- \cap {H_2}^-$

It remains to be shown that it does not always happen that:

$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$

The result is demonstrated by Proof by Counterexample.

Sources

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