Fortissimo Space is not Weakly Countably Compact

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Theorem

Let $T = \struct {S, \tau_p}$ be a Fortissimo space.


Then $T$ is not weakly countably compact.


Proof

It suffices to show that $T$ has an infinite subset without limit points.


Consider the set $S \setminus \set p$.

Let $x \in S$.

We have:

\(\displaystyle \paren {S \setminus \paren {S \setminus \set p} } \cup \set x\) \(=\) \(\displaystyle \set p \cup \set x\)
\(\displaystyle \) \(=\) \(\displaystyle \set {p, x}\)


By definition, $x$ is a limit point of $S \setminus \set p$ if and only if $\set {p, x}$ is not a neighborhood of $x$.


By definition of Fortissimo space, $\set {p, x}$ is open in $T$.

Hence it is a open neighborhood of $x$.

Therefore $x$ is not a limit point of $S \setminus \set p$.


Since $x$ is arbitrary, $S \setminus \set p$ has no limit points.

Hence $T$ is not weakly countably compact.

$\blacksquare$


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