Fortissimo Space is not Sequentially Compact

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Theorem

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


Then $T$ is not sequentially compact.


Proof

Let $T \ ' = \left({S \setminus \left\{{p}\right\}, \tau_p}\right)$ be the subspace of $T$ induced by $S \setminus \left\{{p}\right\}$.

From Discrete Subspace of Fortissimo Space, $T \ '$ is a discrete topological space.


Let $H = \left \langle {x_n}\right \rangle_{n \in \N}$ be a sequence of distinct terms of $S \setminus \left\{{p}\right\}$.

That is:

$\forall n \in \N: x_n \ne p$

and that:

$n \ne m \implies x_n \ne x_m$


By definition, $H$ is a sequence of distinct terms of $S \setminus \left\{{p}\right\}$.

From Convergence of Sequence in Discrete Space: Corollary, $H$ does not converge in $T \ '$.


$H$ is by definition countable, so from Closed Sets of Fortissimo Space it follows that $H$ is closed.

Thus by definition $S \setminus H$ is open.

Also, as $p \notin H$ by definition of $H$, it follows that $P \in S \setminus H$.

By definition, $S \setminus H$ is an open neighborhood of $p$.

Suppose $H$ converges to the limit $p$

Then, by definition of convergence, for any open set $U \in \tau_p$ such that $p \in U$, there exists some $N \in \R$ such that $n > N \implies x_n \in U$.

But $H \cap \left({S \setminus H}\right) = \varnothing$, and so $H$ cannot converge to $p \in S \setminus H$.


The result follows from the fact that every subsequence of $H$ is also a sequence of distinct terms of $S \setminus \left\{{p}\right\}$.

$\blacksquare$


Sources