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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $2$