Fortissimo Space is not Pseudocompact

Theorem

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

Then $T$ is not a pseudocompact space.

Proof

Let $N_p$ be a neighborhood of $p$ such that $\relcomp S {N_p}$ is countable.

Let $\psi: \relcomp S {N_p} \to \Z_{\ne 0}$ be a bijection between $\relcomp S {N_p}$ and the non-zero integers $\Z_{\ne 0}$.

Let $\phi: S \to \set {0, 1}$ be the mapping defined as:

$\forall x \in S: \map \phi x = \begin {cases} 0 & : x \in N_p \\ \map \psi x & : x \notin N_p \end {cases}$

Then $\phi$ is a continuous mapping.

However, $\phi$ is not bounded.

Hence the result by definition of pseudocompact space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $3$