Closure of Subset of Closed Set of Topological Space is Subset/Proof 2
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Theorem
Let $T$ = $\struct {S, \tau}$ be a topological space.
Let $F$ be a closed set of $T$.
Let $H \subseteq F$ be a subset of $F$.
Let $H^-$ denote the closure of $H$.
Then $H^- \subseteq F$.
Proof
Let $x \notin F$.
Then $x$ is in the open set $S \setminus F$.
From Set Difference with Subset is Superset of Set Difference:
- $S \setminus F \subseteq S \setminus H$
From Subsets of Disjoint Sets are Disjoint:
- $S \setminus F \cap H = \O$
From Set is Open iff Neighborhood of all its Points:
- $S \setminus F$ is an neighborhood of $x$
By definition of the closure of $H$:
- $x \notin H^-$
We have shown that:
- $S \setminus F \subseteq S \setminus H^-$
From Set Difference with Subset is Superset of Set Difference:
- $H^- \subseteq F$
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous): $\S 4.6$: Closure, Interior, Boundary: Lemma $4.3$