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

From ProofWiki
Jump to navigation Jump to search

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 $x$


Then $\struct{S, \tau}$ is a sober space if and only if:

for each completely prime filter $\FF$ in the complete lattice $\struct{\tau, \subseteq}$:
$\exists ! x \in S : \FF = \map \UU x$


Proof

Recall by definition of system of open neighborhoods:

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

Necessary Condition

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


Let $\FF$ be a completely prime filter of $\struct{\tau, \subseteq}$.


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


By the contrapositive statement of the definition of completely prime filter:

$W \notin \FF$


From Completely Prime Filter Induces Meet-Irreducible Open Set:

$W$ is a meet-irreducible open set

By definition of sober space:

$\exists ! x \in S : W = S \setminus \set x^-$


We have:

\(\ds \forall U \in \tau: \, \) \(\ds U \notin \FF\) \(\leadsto\) \(\ds U \subseteq W\) Set is Subset of Union
\(\ds \) \(\leadsto\) \(\ds U \subseteq S \setminus \set x^-\) As $W = S \setminus x^-$
\(\ds \) \(\leadstoandfrom\) \(\ds U \notin \map \UU x\) Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure


We also have

\(\ds \forall U \in \tau: \, \) \(\ds U \notin \map \UU x\) \(\leadsto\) \(\ds U \subseteq S \setminus \set x^-\) Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure
\(\ds \) \(\leadsto\) \(\ds U \subseteq W\) As $W = S \setminus x^-$
\(\ds \) \(\leadsto\) \(\ds U \notin \FF\) Filter Axiom $\paren{3}$ and $W \notin \FF$


Hence:

$\forall U \in \tau : U \notin \FF \iff U \notin \map \UU x$


From Biconditional Equivalent to Biconditional of Negations:

$\forall U \in \tau : U \in \FF \iff U \in \map \UU x$

By definition of set equality:

$\FF = \map \UU x$


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

$x$ is unique


The result follows.

$\Box$


Sufficient Condition

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


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sober_Space_iff_Completely_Prime_Filter_is_Unique_System_of_Open_Neighborhoods&oldid=687405"