# Closure of Intersection may not equal Intersection of Closures

## 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:

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

## Outline of 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.

## Proof 1

Let $\struct {\R, \tau}$ be the real number line under the usual (Euclidean) topology.

Let $\Q$ denote the set of rational numbers.

Let $\R \setminus \Q$ denote the set of irrational numbers.

$\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$
$\Q^- \cap \paren {\R \setminus \Q}^- = \R$

The result follows.

$\blacksquare$

## Proof 2

Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.

Let $H_1 = \openint 0 {\dfrac 1 2}$ and $H_2 = \openint {\dfrac 1 2} 1$.

By inspection it can be seen that:

$H_1 \cap H_2 = \O$

Thus from Closure of Empty Set is Empty Set:

$\paren {H_1 \cap H_2}^- = \O$
$H_1 = \closedint 0 {\dfrac 1 2}, H_2 = \closedint {\dfrac 1 2} 1$

Thus:

${H_1}^- \cap {H_2}^- = \set {\dfrac 1 2}$

So $\paren {H_1 \cap H_2}^- \ne {H_1}^- \cap {H_2}^-$

Hence the result.

$\blacksquare$

## Examples

### Arbitrary Subsets of $\R$

Let $H$ and $K$ be subsets of the set of real numbers $\R$ defined as:

 $\ds H$ $=$ $\ds \openint 0 2 \cup \openint 3 4$ $\ds K$ $=$ $\ds \openint 1 3$

Let $\map \cl H$ denote the closure of $H$.

Then:

$H \cap \map \cl K$
$\map \cl H \cap K$
$\map \cl H \cap \map \cl K$
$\map \cl {H \cap K}$

are all different.