Topological Closure of Subset is Subset of Topological Closure/Proof 2

From ProofWiki
Jump to navigation Jump to search


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

Let $H \subseteq K$ and $K \subseteq S$.


$\map \cl H \subseteq \map \cl K$

where $\map \cl H$ denotes the closure of $H$.


From the definition of closure:

$\map \cl H$ is the union of $H$ and its limit points.

Let $x \in \map \cl H$.

If $x \in H$ then $x \in K \implies x \in \map \cl K$.

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 $x \in \map \cl K$.

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.