# Underlying Set of Topological Space is Closed

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## Theorem

Let $T = \left({S, \tau}\right)$ 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 $T = \left({S, \tau}\right)$ if and only if $S \setminus U$ is closed in $T$.

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

From Set Difference with Empty Set is Self, we have $S \setminus \varnothing = S$, so $S$ is closed in $T$.

$\blacksquare$

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 3.2$: Topological Spaces: Exercise $4$