Empty Set is Closed/Topological Space

Theorem

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


Then $\O$ 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 $T$.

By definition of topological space, $S$ is open in $T$.

From Set Difference with Self is Empty Set:

$S \setminus S = \O$

Hence $\O$ 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 $4$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.