Compact Hausdorff Space with no Isolated Points is Uncountable/Lemma

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Lemma for Compact Hausdorff Space with no Isolated Points is Uncountable

Let $T = \left({S, \tau}\right)$ be a Hausdorff space with no isolated points.

Let $U \subseteq S$ be a non-empty open set in $T$.

Let $x \in S$.


Then there is a non-empty open subset $V \subseteq U$ such that $x \notin V^-$, where $V^-$ is the closure of $V$.


Proof

First we show that there is a $y \in U$ such that $y \ne x$:

By Law of Excluded Middle, either $x \in U$ or $x \notin U$.


Let $x \in U$.

We have that $x$ is not an isolated point.

Therefore:

$\exists y \in U: y \ne x$


Let $x \notin U$.

As $U$ is non-empty it has an element $y$, and $y \ne x$.

Thus in either case, there is a $y \in U$ such that $y \ne x$.

$\Box$


By Closure Condition for Hausdorff Space, there is an open set $V$ such that $y \in V$ and $x \notin V^-$.

$\blacksquare$

Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.