Sober Space iff Completely Prime Filter is Unique System of Open Neighborhoods/Sufficient Condition

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Theorem

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

For each $x \in S$, let:

$\map \UU x$ denote the system of open neighborhoods of $x$

For each completely prime filter $\FF$ in the complete lattice $\struct{\tau, \subseteq}$, let:

$\exists ! x \in S : \FF = \map \UU x$


Then $\struct{S, \tau}$ is a sober space.


Proof

Recall by definition of system of open neighborhoods:

$\forall x \in S : \map \UU x = \set{U \in \tau : x \in U}$


For each completely prime filter $\FF$ in the complete lattice $\struct{\tau, \subseteq}$, let:

$\exists ! x \in S : \FF = \map \UU x$


Let $W \in \tau$ be a meet-irreducible open set.


Let $\FF = \set{U \in \tau : U \nsubseteq W}$.

From Meet-Irreducible Open Set Induces Completely Prime Filter:

$\FF$ is a completely prime filter


We have by hypothesis:

$\exists ! x \in S : \FF = \map \UU x$


We have:

\(\ds S \setminus \set x^-\) \(\in\) \(\ds \tau \setminus \map \UU x\) Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure
\(\ds \leadsto \ \ \) \(\ds S \setminus \set x^-\) \(\in\) \(\ds \tau \setminus \FF\) as $\map \UU x = \FF$
\(\ds \leadsto \ \ \) \(\ds S \setminus \set x^-\) \(\subseteq\) \(\ds W\) Definition of $\FF$


We also have:

\(\ds W\) \(\notin\) \(\ds \FF\) Definition of $\FF$
\(\ds \leadsto \ \ \) \(\ds W\) \(\notin\) \(\ds \map \UU x\) as $\map \UU x = \FF$
\(\ds \leadsto \ \ \) \(\ds W\) \(\in\) \(\ds \tau \setminus \map \UU x\) Definition of Set Difference
\(\ds \leadsto \ \ \) \(\ds W\) \(\subseteq\) \(\ds S \setminus \set x^-\) Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure


By definition of set equality:

$W = S \setminus \set x^-$


From System of Open Neighborhoods are Equal Iff Singleton Closures are Equal:

$x$ is unique


The result follows.

$\blacksquare$

Also see

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