Characterization of T1 Space using Basis

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $\BB$ be a basis for $T$.

Then:

$T$ is a $T_1$ Space

if and only if

$\forall x, y \in S : x \ne y$, both:

$\exists B_x \in \BB : x \in B_x, y \notin B_x$

and:

$\exists B_y \in \BB : y \in B_y, x \notin B_y$

Proof

Necessary Condition

From Basis induces Local Basis:

$\forall x \in S : \BB_x = \set{B \in \BB : x \in B}$ is a local basis of $x$

By definition of local basis:

$\forall x \in S : \BB_x$ is a neighborhood basis of open sets

From Characterization of T1 Space using Neighborhood Basis:

$\forall x, y \in S : x \ne y$, both:

$\exists B_x \in \BB_x : y \notin B_x$

and:

$\exists B_y \in \BB_y : x \notin B_y$

Hence:

$\forall x, y \in S : x \ne y$, both:

$\exists B_x \in \BB : x \in B_x, y \notin B_x$

and:

$\exists B_y \in \BB : y \in B_y, x \notin B_y$

$\Box$

Sufficient Condition

Let $T$ satisfy:

$\forall x, y \in S : x \ne y$, both:

$\exists B_x \in \BB : x \in B_x, y \notin B_x$

and:

$\exists B_y \in \BB : y \in B_y, x \notin B_y$

By definition of basis:

$\BB \subseteq \tau$.

It follows that:

$\forall x, y \in S : x \ne y$, both:

$\exists B_x \in \tau : x \in B_x, y \notin B_x$

and:

$\exists B_y \in \tau : y \in B_y, x \notin B_y$

Hence $T$ is a $T_1$ Space by definition.

$\blacksquare$