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

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

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

Then:
 * $T$ is a $T_1$ Space


 * $\forall x, y \in S : x \ne y$, both:
 * $\exists B_x \in \BB : x \in B_x, y \notin B_x$
 * and:
 * $\exists B_y \in \BB : y \in B_y, x \notin B_y$

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 User:Leigh.Samphier/Topology/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:
 * $\exists B_y \in \BB_y : x \notin B_y$

By definition of $\BB_x$ for all $x \in S$:
 * $\forall x, y \in S : x \ne y$, both:
 * $\exists B_x \in \BB : x \in B_x, y \notin B_x$
 * and:
 * $\exists B_y \in \BB : y \in B_y, x \notin B_y$

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.