Open Neighborhoods of Point form Directed Ordering

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

Let $x \in S$.

Let $\NN \subseteq \tau$ be the set of open neighborhoods of $x$.

Then $\supseteq$, the ordering of $\NN$ by reverse inclusion, is a directed ordering on $\NN$.

Proof
By Subset Relation is Ordering and Dual Ordering is Ordering $\supseteq$ is an ordering on $\NN$.

To show that $\supseteq$ is directed, let $U, V \in \NN$.

Then $x \in U, V$, so that $x \in U \cap V$.

Hence $U \cap V \in \NN$ is an open neighborhood of $x$.

Moreover by Intersection is Subset:


 * $U, V \supseteq U \cap V$

Hence $\supseteq$ is directed.