Interior of Closed Set of Particular Point Space
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Theorem
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $V \subseteq S$ be closed in $T$ such that $V \ne S$.
Then:
- $V^\circ = \O$
where $V^\circ$ denotes the interior of $V$.
Proof
By definition:
- $\forall U \in \tau_p, U \ne \O: p \in U$
Thus if $V$ is closed in $T$:
- $\exists U \subseteq T: V = \relcomp S U$
So $p \notin V$.
Hence no open set of $T$ can be a subset of $V$ unless it is $\O$.
Hence the result, by definition of interior.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $2$