# Convergent Sequences of Extremally Disconnected Hausdorff Spaces

It's written in Willard (15G.3) that the only convergent sequences in a Hausdorff Extremally Disconnected space are the eventually constant sequences. However, it has not provided a proof.

I've tried to derive this myself, but am unable to do so. There's another post on Math Stackexchange about this problem, but a solution wasn't presented there. So, any help in proving this is appreciated!

• It does provide the idea for the proof; it's an exercise, so the reader should supply the proof. – Henno Brandsma Jul 26 '20 at 7:45
• It has, but I've not been able to develop a proof with it. – Ishan Deo Jul 26 '20 at 8:00

Let $$(x_n)$$ be a sequence in $$X$$ such that $$x_n \to p \in X$$.

Assume it is not eventually constant. Then in particular $$x_n \ne p$$ for infinitely many $$n$$, i.e. there exists a subsequence $$(x_{n _k})$$ such that $$y_k = x_{n _k} \ne p$$ for all $$k$$. Let us inductively construct a subsequence $$(y_{k_r})$$ of $$(y_k)$$ and a sequence of pairwise disjoint open sets $$U_r$$ such that $$y_{k_r} \in U_r$$ and $$p \notin \overline U_r$$.

For $$r=1$$ we take $$k_1 = 1$$. There exist disjoint open neigborhoods $$U_1$$ of $$y_1$$ and $$V_1$$ of $$p$$. Thus $$p \notin \overline U_1$$. For the induction step observe that $$W_r = X \setminus \bigcup_{i=1}^r \overline U_i$$ is an open neigborhood of $$p$$. Since $$y_k \to p$$, we find $$k_{r+1}$$ such that $$y_{k_{r+1}} \in W$$. There exist disjoint open subsets $$U_{r+1}$$ and $$V_{r+1}$$ of $$W_r$$ such that $$y_{k_{r+1}} \in U_{r+1}$$ and $$p \in V_{r+1}$$. These subsets are also open in $$X$$. Clearly $$p \notin \overline U_{r+1}$$ and $$U_{r+1} \cap \bigcup_{i=1}^r \overline U_i = \emptyset$$ which shows that $$U_1,\ldots, U_{r+1}$$ are pairwise disjoint.

Let $$U = \bigcup_{i=1}^\infty U_{2i}$$ which is open. Thus also $$\overline U$$ is open. Since $$y_{k_{2i}} \in U$$ and $$y_{k_{2i}} \to p$$, we have $$p \in \overline U$$. Therefore $$y_{k_r} \in \overline U$$ for $$r \ge R$$. Now let $$r \ge R$$ be odd. Then $$U_r \cap U = \emptyset$$, thus $$U_r \cap \overline U = \emptyset$$ because $$U \subset X \setminus U_r$$. We conclude that $$y_{k_r} \notin \overline U$$. This is the desired contradiction.

As an alternative to the Willard sketch of proof that Paul worked out, note that if $$(x_n)_n$$ is a sequence in $$X$$ (a Hausdorff space) that is not eventually constant and does not have a constant subsequence, the set $$A= \{x_n: n \in \Bbb N\}$$ is infinite and so has an infinite pairwise disjoint family of sets. It follows that we have points $$x_{n_k}$$ and pairwise disjoint open subsets $$U_k$$ (containing $$x_{n_k}$$) for all $$k$$, forming a subsequence of $$(x_n)$$.

We switch to the extremally disconnected setting for $$X$$:

Now assume $$x_n \to p$$ for some $$p \in X$$. Then define $$U=\bigcup \overline{U_k}$$, which is open and thus $$C^\ast$$-embedded in $$X$$ (by an earlier part of that same exercise) and define $$f: U \to [0,1]$$ by $$f(x)=0$$ when $$x \in \overline{U_k}$$ for $$k$$ even, and $$=1$$ for $$k$$ odd, which is clearly continuous (we define the function on disjoint open sets). So we can extend this to a continuous $$F: X \to [0,1]$$ but then we have a contradiction as $$F(x_{n_{2k}}) \to 0$$ while $$F(x_{n_{2k+1}}) \to 1$$ while both subsequences should converge to $$F(p)$$, so the sequence cannot converge.