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Let $X$ be a discrete space; consider the space $\beta(X)$.

a) Show that if $A\subset X$, then $\overline{A}$ and $\overline{X-A}$ are disjoint, where the closures are taken in $\beta(X)$.

b) Show that if $U$ is open in $\beta(X)$, then $\overline{U}$ is open in $\beta(X)$.

c) Show that $\beta(X)$ is totally disconnected.

I was able to solve the parts a) and b) but cannot solve the last one.

I would be very grateful if anyone can show to solve it.

RFZ
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  • which definition of $\beta X$ do you work with (ultrafilters, or closure in a certain space of functions)? Note $f:X\to2={0,1},f(A)=0,f(X-A)=1$ could be extended continuously to $\beta X$ and would separate the closures. Wouldn't (c) follow from (b)? I have to think, perhaps Extremally disconnected implies totally disconnected? (It does, I just need to find a suitable reference, see Engelking, or google it.) – Mirko Oct 27 '19 at 19:29
  • @Mirko, Probably (c) follows from (b). But i don't see it. I would like to see the solution since I have spent some time and failed to prove it. I learnt $\beta X$ from Munkres book which describes it as a compact Hausdorff space $\beta X$ containing $X$ as a subspace such that any continuous function $f:X\to C$ where $C$-compact Hausdorff could be extended uniquely to continuous function $g:\beta X\to C$. – RFZ Oct 27 '19 at 19:33
  • Extremally disconnected implies totally disconnected https://stacks.math.columbia.edu/tag/08YH though I am searching for a better reference (or else I would have to think of the actual proof :) – Mirko Oct 27 '19 at 19:38
  • @Mirko, but the link which you gave does not contain a proof – RFZ Oct 27 '19 at 19:40
  • But at least now you could think of proving this more general result, that in Hausdorff spaces extremally disconnected implies totally disconnected (instead of focusing on properties of $\beta X$ ) – Mirko Oct 27 '19 at 19:41
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    just a couple of related links https://www.jstor.org/stable/2035286 also at https://www.ams.org/journals/proc/1967-018-02/S0002-9939-1967-0210066-0/S0002-9939-1967-0210066-0.pdf , and a related MSE question https://math.stackexchange.com/q/632246 – Mirko Oct 27 '19 at 20:11

1 Answers1

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A space is called extremally disconnected if the closure of every open set is open. It is known that extremally disconnected Hausdorff spaces are totally disconnected (every component of connectedness is a singleton). Thus, (c) follows from (b).

Suppose that $E$ is any extremally disconnected Hausdorff space, and take any $x\in E$. We will show that the component of $x$ is $\{x\}$. Take any $y\neq x$, we will show that there is a clopen set containing $x$ but not $y$. Start with disjoint open neighborhoods $U$ of $x$ and $V$ of $y$. Then $\overline U$ is clopen, and it is disjoint from $V$ (since $U\subseteq E\setminus V$ and using that $E\setminus V$ is closed we get that $\overline U\subseteq E\setminus V$). Thus $\overline U$ is a clopen set contaning $x$ but not $y$, showing that $y$ is not in the component of connectedness of $x$.

Mirko
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  • Dear Mirko! You have demonstated a very nice proof. I have tried a lot and failed to do it. However, the proof is not so difficult and it is quite nice! +1! I'll accept your answer as the best! Thanks a lot for your permamnent help! – RFZ Oct 28 '19 at 01:02
  • @ZFR You are welcome! – Mirko Oct 28 '19 at 01:06