Fortissimo Space is Excluded Point Space with Countable Complement Space

Theorem

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

Then $\tau_p$ is the minimal topology that is generated by the excluded point topology and the countable complement topology.

Proof

Let $T_1 = \struct {S, \tau_1}$ be the excluded point space on $S$ from $p$.

Let $T_2 = \struct {S, \tau_2}$ be the countable complement space on $S$.

By definition:

$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$

$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is countable $\rightset{} \cup \set \O$

By definition of Fortissimo space, we have:

$U \in \tau_1 \implies U \in \tau_p$

$U \in \tau_2 \implies U \in \tau_p$

So $\tau_1 \cup \tau_2 \subseteq \tau_p$.

Similarly:

$U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$

and so $\tau_p \subseteq \tau_1 \cup \tau_2$.

So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from Union is Smallest Superset.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $1$