Completely Hausdorff Property is Hereditary

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Let $T_H = \left({H, \tau_H}\right)$, where $\varnothing \subset H \subseteq S$, be a subspace of $T$.


Then $T_H$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.


Proof

Let $T = \left({S, \tau}\right)$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Then:

$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \varnothing$

That is, for any two distinct elements $x, y \in S$ there exist open sets $U, V \in \tau$ containing $x$ and $y$ respectively whose closures are disjoint.


We have that the set $\tau_H$ is defined as:

$\tau_H := \left\{{U \cap H: U \in \tau}\right\}$


Let $x, y \in H$ such that $x \ne y$.

Then as $x, y \in S$ we have that:

$\exists U, V \in \tau: x \in U, y \in V, U^- \cap V^- = \varnothing$

As $x, y \in H$ we have that:

$x \in U \cap H, y \in V \cap H: \left({U^- \cap H}\right) \cap \left({V^- \cap H}\right) = \varnothing$

From Closure in Subspace we have that:

$U^- \cap H = \left({U \cap H}\right)^-$
$V^- \cap H = \left({V \cap H}\right)^-$

So:

$x \in U \cap H, y \in V \cap H: \left({U \cap H}\right)^- \cap \left({V \cap H}\right)^- = \varnothing$

and so the $T_{2 \frac 1 2}$ (completely Hausdorff) axiom is satisfied in $H$.

$\blacksquare$


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