Characterization of T1 Space using Neighborhood Basis

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Theorem

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

For each $x \in S$, let $\NN_x$ be a neighborhood basis at $x$.

Then:

$T$ is a $T_1$ Space

if and only if

$\forall x, y \in S : x \ne y$, both:

$\exists N \in \NN_x : y \notin N$

and:

$\exists M \in \NN_y : x \notin M$

Proof

Necessary Condition

Let $T$ be a $T_1$ Space.

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

By definition of $T_1$ Space:

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

From Set is Open iff Neighborhood of all its Points:

$U$ is a neighborhood of $x$

By definition of neighborhood basis:

$\exists N \in \NN_x : N \subseteq U$

It follows that:

$y \notin N$

Similarly: $\exists M \in \NN_y : x \notin M$

The result follows.

$\Box$

Sufficient Condition

Let $T$ satisfy:

$\forall x, y \in S : x \ne y$, both:

$\exists N \in \NN_x : y \notin N$

and:

$\exists M \in \NN_y : x \notin M$

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

by hypothesis:

$\exists N \in \NN_x : y \notin N$

By definition of neighborhood:

$\exists U \in \tau : x \in U : U \subseteq N$

It follows that:

$x \in U, y \notin U$

Similarly:

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

Since $x, y$ were arbitrary:

$\forall x, y \in S$ such that $x \ne y$, both:

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

and:

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

It follows that $T$ is a $T_1$ Space by definition.

$\blacksquare$