Fort Space is Sequentially Compact
Theorem
Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Then $T$ is a sequentially compact space.
Proof
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $T$.
Suppose $\sequence {x_n}$ takes an infinite number of distinct values in $S$.
Then there is an infinite subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ with distinct terms.
Let $U$ be a neighborhood of $p$.
Then $S \setminus U$ is a finite set by definition.
Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$.
Thus $\sequence {x_{n_r} }$ converges to $p$.
Otherwise $\sequence {x_n}$ only takes a finite number of distinct values.
Then, since $\sequence {x_n}$ is infinite, there exists $x \in S$ such that:
- $\forall N \in \N: \exists n > N: x = x_n$
This implies that we can take a subsequence of $\sequence {x_n}$ which is constant, and which converges to that constant.
We can conclude then that, by definition, $T$ is a sequentially compact space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $4$