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!

  • $\begingroup$ It does provide the idea for the proof; it's an exercise, so the reader should supply the proof. $\endgroup$ Jul 26, 2020 at 7:45
  • $\begingroup$ It has, but I've not been able to develop a proof with it. $\endgroup$
    – Ishan Deo
    Jul 26, 2020 at 8:00
  • $\begingroup$ Note that the hint in the book is faulty: it directly contradicts the fact that $X$ is extremally disconnected. $\endgroup$
    – C Squared
    May 12 at 3:53

2 Answers 2


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.


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