Topological Closure of Subset is Subset of Topological Closure
Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq K$ and $K \subseteq S$.
Then:
- $\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({K}\right)$
where $\operatorname{cl}\left({H}\right)$ denotes the closure of $H$.
Proof 1
From Topological Closure is Closed, $\operatorname{cl} \left({K}\right)$ is closed.
From Set is Subset of its Topological Closure:
- $K \subseteq \operatorname{cl} \left({K}\right)$
By Subset Relation is Transitive, it follows that:
- $H \subseteq \operatorname{cl} \left({K}\right)$
Hence, by definition of closure as the smallest closed set that contains $H$:
- $\operatorname{cl} \left({H}\right) \subseteq \operatorname{cl} \left({K}\right)$
$\blacksquare$
Proof 2
From the definition of closure:
- $\operatorname{cl}\left({H}\right)$ is the union of $H$ and its limit points.
Let $x \in \operatorname{cl} \left({H}\right)$.
If $x \in H$ then $x \in K \implies x \in \operatorname{cl} \left({K}\right)$.
Otherwise $x$ is a limit point of $H$.
That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.
But as $y \in H$ it follows that $y \in K$.
So every open set $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$.
This is the definition for a limit point of $K$.
Thus $x \in \operatorname{cl} \left({K}\right)$.
A part of this page has to be extracted as a theorem: Extract the above paragraph into something like "Limit Point of Subset is Limit Point" (which proves that $H \subseteq K \implies H' \subseteq K'$). Then use Set Union Preserves Subsets to prove this result. |
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3.7$: Definitions: Proposition $3.7.15 \ \text{(b)}$