Interior of Intersection may not equal Intersection of Interiors

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Theorem

Let $T$ be a topological space.


Let $\mathbb H$ be a set of subsets of $T$.

That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.


Then it is not necessarily the case that:

$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$

where $H^\circ$ denotes the interior of $H$.


Proof

From Intersection of Interiors contains Interior of Intersection it is seen that it is always the case that:

$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$

From Interior of Finite Intersection equals Intersection of Interiors it is seen that if $\mathbb H$ is finite, then:

$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ = \bigcap_{H \mathop \in \mathbb H} H^\circ$


It remains to be shown that in the case where $\mathbb H$ is infinite, then it is possible that:

$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \ne \bigcap_{H \mathop \in \mathbb H} H^\circ$


Proof by Counterexample:

Let $A_n$ denote the open real interval:

$A_n := \openint {1 - \dfrac 1 n} {1 + \dfrac 1 n}$

for all $n \in \Z_{>0}$.

From Open Sets in Real Number Line, each of $A_n$ is an open set in $\R$.

From Interior of Open Set:

$\forall n \in \Z_{>0}: A_n^\circ = A_n$

Thus:

$\ds \bigcap A_n = \bigcap A_n^\circ = \set 1$

However:

$\ds \paren {\bigcap A_n}^\circ = \set 1^\circ = \O$

Hence the result.

$\blacksquare$


Also see


Sources

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