Underlying Set of Topological Space is Closed
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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 $4$