Topological Space is Discrete iff All Points are Isolated

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.


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


Proof

Necessary Condition

Let $T = \left({S, \tau}\right)$ be the discrete space on $S$.

Then by definition $\tau = \mathcal P \left({S}\right)$, that is, $\tau$ is the power set of $S$.


Let $x \in S$.

Then from Set in Discrete Topology is Clopen it follows that $\left\{{x}\right\}$ is open in $T$.

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

$\Box$


Sufficient Condition

Suppose that $T = \left({S, \tau}\right)$ is not the discrete space on $S$.

Then from Basis for Discrete Topology it follows that:

$\mathcal B := \left\{{\left\{{x}\right\}: x \in S}\right\}$

is not a basis for $T$.

Thus:

$\exists x \in S: \left\{{x}\right\} \notin \tau$

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

Then as $U \ne \left\{{x}\right\}$ 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.

$\Box$


Hence the result.

$\blacksquare$


Sources