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:
\(\ds \paren {S \setminus \paren {S \setminus \set p} } \cup \set x\) | \(=\) | \(\ds \set p \cup \set x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {p, x}\) |
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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $2$