Arens-Fort Space is Not Extremally Disconnected

From ProofWiki
Jump to: navigation, search

Theorem

Let $T = \left({S, \tau}\right)$ be the Arens-Fort space.


Then $T$ is not extremally disconnected.


Proof

Let $S_m \left({V}\right)$ denote the set $S_m(V) := \left\{{n: \left({m, n}\right) \notin V}\right\}$ where $V \subseteq \Z_{\ge 0} \times \Z_{\ge 0}$ (the same set $S_m$ used in the definition of the Arens-Fort space).

Let $U = \left\{{\left({n, m}\right): \exists k: m = 2k}\right\} \setminus \left\{{\left({0, 0}\right)}\right\}$.

From the definition of the Arens-Fort space, $U$ is open in $T$ since $\left({0, 0}\right) \notin U$.

We have that:

$\left({0, 0}\right) \in \complement_S \left({U}\right)$

and:

$S_m \left({U}\right)$ is infinite for any $m \in \N$.

Thus by definition of the Arens-Fort space, $\complement_S \left({U}\right)$ is not open in $T$.

So $U$ is not closed in $T$.


Let $U^-$ denote the closure of $U$.

We have that:

$\left({0, 0}\right) \notin \complement_S \left({U \cup \left\{{\left({0, 0}\right)}\right\}}\right)$

Therefore by definition of the Arens-Fort space, $\complement_S \left({U \cup \left\{{\left({0, 0}\right)}\right\}}\right)$ is open in $T$.

Therefore $U \cup \left\{{\left({0, 0}\right)}\right\}$ is closed in $T$.

From Set Closure is Smallest Closed Set, it follows that $U^- = U \cup \left\{{\left({0, 0}\right)}\right\}$.


We have that:

$\left({0, 0}\right) \in U \cup \left\{{\left({0, 0}\right)}\right\}$

and:

$S_m \left({U \cup \left\{{\left({0, 0}\right)}\right\}}\right)$ is infinite for any $m \in \N$.

Therefore by definition of the Arens-Fort space, $U \cup \left\{{\left({0, 0}\right)}\right\}$ is not open.


Thus we have created an open set $U$ whose closure $U^-$ is not itself open.

Thus by definition the Arens-Fort space is not extremally disconnected.

$\blacksquare$

Sources

Personal tools
Namespaces
Variants
Actions
Navigation
ProofWiki.org
ToDo
Toolbox
Google AdSense