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Is there a boolean space $X$ without isolated points with the property that no point $x\in X$ is the limit of a long sequence $(x_i)_{i\in I}$ from $X\setminus \lbrace x\rbrace $ ('long sequence' here means that $I$ is any totally ordered index set).

I assume that there is such a space and searched the book "Counterexamples in topology", but could not spot it.

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It seems that there is no such space. Indeed, I claim that every non-isolated point in a Boolean space is the limit of a long sequence in that space.

We may assume that the space $X$ is the Stone space of a Boolean algebra $B$, so that the points in $X$ consist of all the ultrafilters in $B$, and the basic open sets have the form $\{U\in X\mid b\in U\}$, where $b\in B$.

I claim that every non-isolated point $U$ in $X$ is a limit of a long sequence. To see this, let $\langle b_\alpha\mid\alpha\lt\delta\rangle$ be a well-ordered enumeration of a minimal sequence of generators of $U$ as an ultrafilter in $B$. That is, $U$ is the filter generated by the $b_\alpha$'s, but it is not generated by any initial segment of this sequence. Since $U$ is not isolated, it follows that $\delta$ must be infinite, and by rearranging if necessary, we may assume that $\delta$ is a limit ordinal. For each $\gamma\lt\delta$, let $F_\gamma$ be the filter generated by $\{b_\alpha\mid\alpha\lt\gamma\}$. Since this initial segment of the generators does not generate all of $U$, we may extend this filter $F_\gamma$ to an ultrafilter $U_\gamma\in X$ distinct from $U$.

I claim that the long sequence $\langle U_\gamma\mid\gamma\lt\delta\rangle$ converges to $U$. Every basic open neighborhood of $U$ is determined by some $b\in U$, which must be in some $F_\beta$ for some $\beta\lt\delta$. It follows for $\beta\leq\gamma\lt\delta$ that we have $b\in F_\gamma\subset U_\gamma$, and so all such $U_\gamma$ are also in the basic open set determined by $b$. In other words, for any basic open set containing $U$, there is a tail of the sequence $U_\gamma$ that is inside it. So $U_\gamma\to U$, as desired.

The length of the sequence converging to $U$ is the same as the cardinality of the minimal generating set of $U$ as a filter.

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Essentially the same argument works when $X$ is any compact Hausdorff space. Just take a minimal sequence of closed neighborhoods of $U$ that generates the filter of all neighborhoods of $U$. By compactness, the intersection any initial segment of this must contain some point other than $U$. –  Eric Wofsey Dec 23 '12 at 12:21
    
@Joel Great (and kind of unexpected). Thanks. –  Marcus Dec 23 '12 at 13:10
1  
Eric, that is very nice! –  Joel David Hamkins Dec 23 '12 at 13:39

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