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My question is: Is my proof of the proposition from Van Douwen´s paper called Characterization of $\beta \mathbb{Q}$ and $\beta \mathbb{R}$ correct?

Claim: $\alpha \mathbb{R}$, the two-point compactification of $\mathbb{R}$ and $\beta \mathbb{R}$ are the only three topological compactifications of $\mathbb{R}.$

($\beta \mathbb{R}$ = the Stone-Čech compactification, $\alpha \mathbb{R}$ = the one-point compactification).

By topological compactification he means such compactification that every automorphism of the original space extends to automorphism of the compactification.

Proof:

This follows directly from the result about the halfline. (The set of all topological compactifications of $\mathbb{H}$ consists of just two elements: $\alpha \mathbb{H}$ and $\beta \mathbb{H}$.)

If we take just the interval $[0, \infty) \subset \mathbb{R}$ and make a closure, we obtain a topological compactification of $[0, \infty)$. This can be done analogically with the interval $(-\infty, 0]$. Both topological compactifications obtained this way have to be of the same "type" (either both are one-point compactifications, or both are Stone-Čech compactifications). That is because for any $x \in \mathbb{R}$, the map $f: x \rightarrow - x$ induces an autohomeomorphism of $\beta \mathbb{R}$. which implies that $\beta[0,\infty)$ is identical with $\beta(-\infty,0]$.

The paper is called "Characterizations of βQ and βR." Van Douwen only writes that the claim above is direct implication of the claim about halfline, but I tried to write it in detail.

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2 Answers 2

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For a compactification $\gamma X$ of a topological space $X$, let $\gamma_r X= \gamma X\setminus X$ denote the residual set of the compactification.

I leave a proof of the first lemma to you:

Lemma 1. If $\gamma X$ is a compactification of a 2nd countable space $X$, then each $\xi\in \gamma_r X$ is the limit of a sequence in $X$.

From now on, I will be working only with $X={\mathbb R}$. Set $X_-=(-\infty, 0]$ and $X_+=[0,\infty)$. Suppose that $\gamma X$ is a compactification of $X$. I let $\gamma_r X_\pm$ denote the intersections of closures of $X_\pm$ (in $\gamma X$) with $\gamma_r X$. Note that $\gamma_r X= \gamma_r(X_-) \cup \gamma_r(X_+)$.

Lemma 2. Suppose that $\nu X_+$ is a topological compactification of $X_+$.
The group of self-homeomorphisms of $X_+$ acts transitively on $\nu_r X_+$.

Proof. Every $\xi\in \nu_r X_+$ is the limit of a strictly increasing sequence in $X_+\setminus \{0\}$. Given $\xi, \xi'\in \nu_r X_+$, let $(x_n), (x_n')$ denote the corresponding sequences. There exists an increasing bijection $$ f: \{0\}\cup Y=\{x_n: n\in {\mathbb N}\}\to Y'=\{0\}\cup \{x'_n: n\in {\mathbb N}\}.$$ Now, extend this bijection linearly to each interval in $X_+\setminus Y$. The resulting map $F: X_+\to X_+$ is a homeomorphism. Since the compactification was assumed to be topological, the homeomorphism $F$ extends to a homeomorphism $F: \nu X_+\to \nu X_+$. By the construction, $F(\xi)=\xi'$. qed

The next lemma is the key to the proof:

Lemma 3. Suppose that $\nu X$ is a topological compactification of $X$ which is not a 1-point compactification. Then:

(i) Both $X_\pm \to X_\pm \cup \nu_r(X_\pm)$ are topological compactifications.

(ii) $\nu_r(X_-)$ is homeomorphic to $\nu_r(X_+)$.

(iii) $\nu_r(X_-), \nu_r(X_+)$ are disjoint in $\nu X$, i.e. $\nu_r X= \nu_r(X_-)\sqcup \nu_r(X_+)$.

Proof. (i) This follows from the assumption that $\nu X$ is a topological compactification of $X$ and the fact that every self-homeomorphism $f_\pm: X_\pm\to X_\pm$ extends to a self-homeomorphism of $X$ (say, by the identity of the complement).

(ii) This was already established in the attempted proof, by considering the homeomorphism $x\mapsto -x$ of $X$.

(iii) If $\nu X$ is the 2-point compactification, there is nothing to be proven. Hence, I will assume that it is not a 2-point compactification, implying that each $\nu_r(X_\pm)$ are not singletons (of course, both are actually infinite). Consider distinct points $\eta, \xi_+\in \nu_r(X_+)\subset \nu X$ and a point $\xi_-\in \nu_r(X_-)\subset \nu X$. By Lemma 2, there exists a homeomorphism $f: X_+\to X_+$ sending $\xi_+$ to $\eta$. Extend $f$ by the identity to $X_-$ and, thus, obtain a homeomorphism $f: X\to X$. Since $\nu X$ is assumed to be a topological compactification of $X$, $f$ extends to a self-homeomorphism $F$ of the compactification $\nu X$. By the construction, $F$ fixes $\xi_-$ and sends $\xi_+$ to $\eta\ne \xi_-$. This implies that $\xi_-\ne \xi_+$ in $\nu X$. qed

Now, back to the original question.

Proposition. Every topological compactification $\nu X$ of $X={\mathbb R}$ is naturally homeomorphic either to the 1-point compactification, or to the 2-point compactification, or to $\beta X$.

Proof. As noted before,$\nu_r(X_-)$ is homeomorphic to $\nu_r(X_+)$. Hence, both compactifications are either 1-points compactifications or Chech-Stone compactifications of $X_\pm$. In the former case, $\nu X$ is either the 1-point or the 2-point compactification of $X$. Thus, I will assume that both compactifications of $X_\pm$ are the Chech-Stone compactifications. Lemma 3(i) implies that both $\nu_r X_\pm$, $\beta_r X_\pm$ are naturally homeomorphic to the residual sets of the Chech-Stone compactifications of $X_\pm$. Let $h_\pm: \beta_r X_\pm \to \nu_r X_\pm$ be these homeomorphisms, extending the identity maps $X_\pm \to X_\pm$. Since $\beta_r X= \beta_r X_- \sqcup \beta_r X_+$ and $\nu_r X= \nu_r X_- \sqcup \nu_r X_+$, the maps $h_\pm$ combine to a bijective continuous map $h: \beta X\to \nu X$ extending the identity map $X\to X$. Reversing the roles of $\beta X, \nu X$, we conclude that $h$ is a homeomorphism. (Alternatively, use that both compactifications are Hausdorff.) \qed

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  • $\begingroup$ Thank you so much for your time! I don´t understand a few points... 1) "... $f$ extends to a self-homeomorphism of $X$ (say, by the identity of the complement)." What do you mean by the note in brackets? How exactly can we be sure that $f$ extends? 2) What do you mean by "naturally homeomorphic"? Is it any special term to be defined? Or is it the same as "homeomorphic"? $\endgroup$ Jul 18, 2022 at 14:39
  • $\begingroup$ Naturally homeomorphuc means "continuously extending the identity map." Do you know how to check continuity of piecewise-continuous maps? $\endgroup$ Jul 18, 2022 at 16:13
  • $\begingroup$ Yes, thank you. Also, by " $\sqcup$ ", do you mean a disjoint union, or it is the same union as the " $\cup$" and just a typo/different notation? $\endgroup$ Jul 18, 2022 at 19:20
  • $\begingroup$ @TerezaTizkova: Yes, this is the standard notation for the disjoint union. $\endgroup$ Jul 18, 2022 at 19:23
  • $\begingroup$ Thanks. Also, at the end, the $\nu_r X = \nu_r X_- \sqcup \nu_r X_+$ is true only with the assumption that these are NOT 1-point compactification remainders? (As in the lemma 3?) $\endgroup$ Jul 18, 2022 at 23:48
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I tried to put together my own answer, so please correct me if I am wrong or write your own.

This follows directly from the result about the halfline - the set of all topological compactifications of $\mathbb{H}$ consists of just two elements: $\alpha \mathbb{H}$ and $\beta \mathbb{H}$. If we take just the interval $[0, \infty) \subset \mathbb{R}$ and make a closure, we obtain a topological compactification of $[0, \infty)$. This can be done analogically with the interval $(-\infty, 0]$. Both topological compactifications obtained this way have to be of the same "type" (either both are one-point compactifications, or both are Stone-Čech compactifications). That is because for any $x \in \mathbb{R}$, the map $f: x \rightarrow - x$ induces an autohomeomorphism of $\beta \mathbb{R}$ which implies that $\beta[0,\infty)$ is identical with $\beta(-\infty,0]$.

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  • $\begingroup$ No, this argument does not prove the desired claim (which, I think, is actually false). $\endgroup$ Jul 15, 2022 at 13:01
  • $\begingroup$ @MoisheKohan The claim is not false, why do you think so? It is known result that $\mathbb{R}$ has this three (Hausdorff) topological compactifications. Also, I am maybe struggling to long with this problem, so that´s why I posted the answer. Will appreciate any corrections to my reasoning. $\endgroup$ Jul 15, 2022 at 13:10
  • $\begingroup$ @MoisheKohan Ah, thank you, you are right. So I am still stuck with this. $\endgroup$ Jul 16, 2022 at 13:24
  • $\begingroup$ The mistake in your argument is in the last sentence. You need to prove the following regarding the residual set of a topological compactification $\gamma {\mathbb R}$: It is either a singleton or a disjoint union of residual sets of two topological compactifications of ${\mathbb R}_{\pm}$. $\endgroup$ Jul 16, 2022 at 15:31
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    $\begingroup$ I will post a proof when I have time... $\endgroup$ Jul 16, 2022 at 16:50

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