Underlying Set of Topological Space is Closed

Theorem

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

Then the underlying set $S$ of $T$ is closed in $T$.

Proof

From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ if and only if $S \setminus U$ is closed in $T$.

From Empty Set is Element of Topology, $\O$ is open in $T$.

From Set Difference with Empty Set is Self:

$S \setminus \O = S$

Hence $S$ is closed in $T$.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $4$