Open Neighborhoods of Point form Directed Ordering
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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.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): Appendix $\text{A}$ Preliminaries: $\S 2.$ Topology