Fortissimo Space is not Compact

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Theorem

Let $T = \left({S, \tau}\right)$ be a Fortissimo space.


Then $T$ is not a compact space.


Proof

From the definition, $T$ is a compact space if every open cover of $T$ has a finite subcover.


Let $N \subset S$ be a countably infinite set such that $p \notin N$.

Then $\complement_S \left({N}\right)$ is open in $T$ by the definition of the Fortissimo space.

Let $q \in S$ such that $q \ne p$.

As $p \in \complement_S \left({\left\{{q}\right\}}\right)$ it follows that $\left\{{q}\right\}$ is open in $T$.

Then:

$\displaystyle S = \complement_S \left({N}\right) \cup \left({\bigcup_{q \in N} \left\{{q}\right\} }\right)$

Let $\mathcal C$ be defined as:

$\mathcal C := \left\{{\left\{{q}\right\}: q \in N}\right\} \cup \left\{{\complement_S \left({N}\right)}\right\}$

Then from the above, $\mathcal C$ is an open cover of $S$.

But all the sets of $\mathcal C$ are pairwise disjoint.

Therefore $\mathcal C$ can have no finite subcover.

Hence the space cannot be compact.

$\blacksquare$


Sources