Closure preserves subset ordering

up:: Closure (Topology)

Let $(X,\tau )$ be a Topological Space, and $U\subseteq V\subseteq X$, and let $\overline{U},\overline{V}$ be their closures.

Since $U\subseteq V$, it is “smaller” than $V$, and thus it is expected that the amount of closed sets which contain it will be higher to/the same as the amount which enclose $V$.

We can see it as follows: let ${\mathcal{F}}_{U,V}$ be the collection of closed sets which enclose $U,V$. Then we have that

$${\mathcal{F}}_U={\mathcal{F}}_V\cup \{F\in \mathcal{F}(\tau )\mid (U\subseteq F)\wedge (V⊈F)\}$$

Since we expect ${\mathcal{F}}_U\supseteq {\mathcal{F}}_V$, we prove that

$$\overline{U}=\bigcap _{{\mathcal{F}}_U}F\subseteq \bigcap _{{\mathcal{F}}_V}F=\overline{V}$$

Thus, closures preserve subset ordering.