User:Leigh.Samphier/Topology/Characterization of T1 Space using Neighborhood Basis

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


 * $\forall x, y \in S : x \ne y$, both:
 * $\exists N \in \NN_x : y \notin N$
 * and:
 * $\exists M \in \NN_y : x \notin M$

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.

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 $x \ne y$.


 * $\exists N \in \NN_x : y \notin N$
 * $\exists N \in \NN_x : y \notin N$

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.