Closure of Intersection may not equal Intersection of Closures/Proof 1

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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

Outline

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.


Counterexample

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.


From Closure of Intersection of Rationals and Irrationals is Empty Set:

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

From Intersection of Closures of Rationals and Irrationals is Reals:

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

The result follows.

$\blacksquare$


Sources

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