Interior preserves subset ordering

up:: Interior (Topology)

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

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

We can see it as follows: let ${\mathcal{T}}_{U,V}$ be the collection of open sets which are contained inside $U,V$. Then we have that

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

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

$$\mathring{U}=\bigcup _{{\mathcal{T}}_U}F\subseteq \bigcup _{{\mathcal{T}}_V}F=\mathring{V}$$

Thus, interiors preserve subset ordering.