Sober Space iff Completely Prime Filter is Unique System of Open Neighborhoods/Necessary Condition
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Theorem
Let $\struct{S, \tau}$ be a $\struct{S, \tau}$ is a sober space.
For each $x \in S$, let:
- $\map \UU x$ denote the system of open neighborhoods of $x$
Then:
- 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:
- $\forall x \in S : \map \UU x = \set{U \in \tau : x \in U}$
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.
$\blacksquare$