Completely Hausdorff Property is Hereditary

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.


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


Proof

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

Then:

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

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:

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


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

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

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

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

$x \in U \cap H, y \in V \cap H: \paren {U^- \cap H} \cap \paren {V^- \cap H} = \O$

From Corollary to Closure in Subspace we have that:

$\paren {U \cap H}^- \subseteq U^- \cap H$
$\paren {V \cap H}^- \subseteq V^- \cap H$

where $\paren{U \cap H}^-$ and $\paren{V \cap H}^-$ are the closures in $H$ of the respective sets.


So:

$x \in U \cap H, y \in V \cap H: \paren {U \cap H}^- \cap \paren {V \cap H}^- = \O$

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

$\blacksquare$


Sources