# Topological Space is Discrete iff All Points are Isolated

## Theorem

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

$\tau$ is the discrete topology on $S$ if and only if all points in $S$ are isolated points of $T$.

## Proof

### Necessary Condition

Let $T = \struct {S, \tau}$ be the discrete space on $S$.

Then by definition $\tau = \powerset S$, that is, $\tau$ is the power set of $S$.

Let $x \in S$.

Then from Set in Discrete Topology is Clopen it follows that $\set x$ is open in $T$.

Thus by definition $x \in S$ is an isolated point of $T$.

$\Box$

### Sufficient Condition

Let $T$ be such that $x \in S$ is an isolated point of $T$ for all $x \in T$.

Aiming for a contradiction, suppose $T = \struct {S, \tau}$ is not the discrete space on $S$.

Then from Basis for Discrete Topology it follows that:

$\BB := \set {\set x: x \in S}$

is not a basis for $T$.

Thus:

$\exists x \in S: \set x \notin \tau$

Let $U \in \tau$ such that $x \in U$.

Then as $U \ne \set x$ it follows that $\exists y \in U: y \ne x$.

That is, there are no open sets of $T$ which contain only $x$.

So $x$ is not an isolated point.

It follows by Proof by Contradiction that $T$ is the discrete space on $S$.

$\Box$

Hence the result.

$\blacksquare$