I am solving Section38 Exercise 5 in Topology, Munkres.

I solved that there is continuous surjectice closed $$f : \beta(S_\Omega) \rightarrow Y$$ for any compactification $Y$ of $S_\Omega$

And one point compactification of $S_\Omega$ is equivalent to Stone-Cech compactification.

However I am stuck in the last problem that Every compactification of $S_\Omega$ is equivalent to one point compactification.

Could you help me with details??

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    $\begingroup$ Don't assume every notation is standard or widely known. What is $S_{\Omega}$? $\endgroup$ – DanielWainfleet Oct 20 '19 at 8:39
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    $\begingroup$ @DanielWainfleet $\omega_1$ in the order topology. Munkres has some weird notation. He denotes $\omega_1 + 1$ by $\overline{S_\Omega}$. $\endgroup$ – Henno Brandsma Oct 20 '19 at 9:06
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    $\begingroup$ Let $S$ be a non-compact Tychonoff space. If $id_S:S\to cS$ is a compactification of $S$ such that $(cS)\setminus S$ has at least $2$ members then $S$ has a disjoint pair $A,B$ of non-compact closed subsets. Equivalently, if at least one of $A,B$ is compact whenever $A,B$ are disjoint closed subsets of $S,$ then $\beta S=\alpha S.$ $\endgroup$ – DanielWainfleet Oct 20 '19 at 9:14
  • $\begingroup$ Since I dk what $S_{\Omega}$ is, I will add: Let $S$ be a normal space and let $id_S:S\to \beta S$ be the Stone-Cech compactification of $S$. If A,B are disjoint closed subsets of $S$ then $A,B$ have disjoint closures in $\beta S. $ Hence for a normal space $S,$ we have $\beta S \alpha S$ iff at least one of $A,B$ is compact whenever $A,B$ are disjoint closed subsets of $S.$ $\endgroup$ – DanielWainfleet Oct 20 '19 at 9:27
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    $\begingroup$ @DanielWainfleet that characterisation of $\beta S$ is not treated in Munkres. Just the extension of bounded real-valued functions one. $\endgroup$ – Henno Brandsma Oct 20 '19 at 9:34

Let $S$ be a Tychonoff space and let $id_S:S\to cS$ be a compactification of $S.$

(i). If $p\in cS \setminus S$ and $U$ is an open subset of $cS$ with $p\in U$ then the closure of $S\cap U$ in $S$ is not compact.

Proof: Suppose otherwise. Then $Cl_S(S\cap U)$ is compact so it is closed in the compact Hausdorff space $cS.$ So $W=U\cap (cS \setminus Cl_S(S\cap U))$ is open in $cS,$ and $W$ is not empty, as $p\in W$. But this contradicts the denseness of $S$ in $cS$ because $$S\cap W= (S\cap U)\cap (S\setminus Cl_S(S\cap U))\subset (S\cap U)\cap (S\setminus (S\cap U))=\emptyset.$$

(ii). If there exist $p,q \in cS\setminus S$ with $p\ne q$ then there are disjoint closed non-compact $A, B$ in $S.$

Proof: Let the closure bar denote closure in $cS.$ Let $U,V$ be open in $cS$ with $p\in U, q\in V,$ and $\bar U\cap \bar V=\emptyset.$ Let $A=Cl_S(S\cap U)$ and $B=Cl_S(S\cap V).$ By (i), neither $A$ nor $B$ is compact. And $$A\cap B=(S\cap \overline {S\cap U})\cap (S\cap \overline {S\cap V})\subset\bar U\cap \bar V=\emptyset.$$

Now $S_{\Omega}$ does not have a disjoint pair of closed non-compact subsets. [...If $A, B$ are closed and non-compact in $S_{\Omega}$ then $A,B$ are closed and unbounded above and so $A\cap B$ is also unbounded above...]. And $S_{\Omega}$ is not compact. So any compactification of $S_{\Omega}$ is equivalent to a one-point compactification.

  • $\begingroup$ When I was first studying compactifications, I managed to obtain (i) somewhat indirectly as a corollary to: Let $Y$ be a dense subspace of $X$ and let $C\subset Y$ such that $C$ is open in $Y$ and $Cl_Y(C)$ is closed in $X$. Then $C$ is open in $X$ and $Cl_Y(C)=Cl_X(C)$, which I "discovered". $\endgroup$ – DanielWainfleet Oct 20 '19 at 10:59
  • $\begingroup$ Thanks a lot. This gives me another critical perspective to understand notion of compactification. $\endgroup$ – HooMun Oct 20 '19 at 11:33
  • $\begingroup$ @HooMun This answer doesn't really match with Munkres theory treatment. It is true though, and it might help you. You do need to show the lemma that any two closed non-compact subsets of $S_\Omega$ intersect. $\endgroup$ – Henno Brandsma Oct 20 '19 at 12:08

In part a) of that exercise it is shown that any continuous function $f: S_\Omega \to \Bbb R$ is eventually constant, in the sense that there is some $\alpha_0 \in S_\Omega$ and some $p \in \Bbb R$ such that $\forall \alpha \ge \alpha_0: f(\alpha) = p$. (In particular all continuous real-valued functions are bounded.)

This means that in particular the inclusion of $S_\Omega$ into $\overline{S}_\Omega$ obeys the extension property: every continuous $f: S_\Omega \to \Bbb R$ has a continuous extension $\bar{f}$ to $\overline{S}_\Omega = S_\Omega \cup \{\Omega\}$: we just give $\bar{f}$ the value $p$ on $\Omega$ too, and as basic neighbourhoods of $\Omega$ in $\overline{S}_\Omega$ are of the form $(\alpha, \Omega]$ (as it's the maximal element), and so the extension is continuous (still constant on almost all basic neighbourhoods).

Now theorem 38.5 essentially says that $\overline{S}_\Omega$ is equivalent to $\beta S_\Omega$. And as $\overline{S}_\Omega\setminus S_\Omega$ has one point, $\overline{S}_\Omega$ is also equivalent to the one-point compactification of $S_\Omega$.

That's all there is to it: the one-point compactification $\overline{S}_\Omega$ obeys the extension property so it's essentially the Cech-Stone compactification.

And if $h: S_\Omega \to C$ is any compactification (so $h\restriction S_\Omega$ is an embedding into a compact Hausdorff $C$ where $h[S_\Omega]$ is dense in $C$) then Thm. 38.4 in that paragraph says that $h$ has a continuous extension $\beta h$ from $\overline{S}_\Omega$ to $C$ and then $C = \beta h[\overline{S}_\Omega]$ by density and continuity and $\beta h$ is a homeomorphism between $\overline{S}_\Omega$ and $C$. (a 1-1 continuous map from a compact space onto a Hausdorff one.). So all compactifications of $S_\Omega$ are just $\overline{S}_\Omega$ in essence.

  • $\begingroup$ What is $S_{\Omega}$? $\endgroup$ – DanielWainfleet Oct 20 '19 at 9:28
  • $\begingroup$ @DanielWainfleet $\omega_1$, in the order topology. The set space $S_\Omega \cup \{\Omega\}$ is denoted $\overline{S}_\Omega$, i.e. $\omega_1 + 1$. It's idiosyncratic, but the OP followed his text and so do I. $\Omega$ is the first uncountable ordinal and he uses the $S$-subcript notation to denote the set with the order topology ($S$ for space, I think). $\endgroup$ – Henno Brandsma Oct 20 '19 at 9:31
  • $\begingroup$ Thank you. I asked the OP but your A appeared so I asked you also.......+1 $\endgroup$ – DanielWainfleet Oct 20 '19 at 9:33
  • $\begingroup$ Actually I understand all ideas above. But I have difficulty to link one-point compactification and any compactification of $S_\Omega$. $\endgroup$ – HooMun Oct 20 '19 at 11:37
  • $\begingroup$ @HooMun Any compactification of $S_\Omega$ has to add at least a point and $\beta S_\Omega = \overline{S}_\Omega$ is the maximal one (as Munkres shows), so it's the only one. $\endgroup$ – Henno Brandsma Oct 20 '19 at 11:45

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