# Proof that $\alpha \mathbb{R}$, the two-point compactification and $\beta \mathbb{R}$ are the only three topological compactifications of $\mathbb{R}$

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.

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

• 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"? Jul 18, 2022 at 14:39
• Naturally homeomorphuc means "continuously extending the identity map." Do you know how to check continuity of piecewise-continuous maps? Jul 18, 2022 at 16:13
• 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? Jul 18, 2022 at 19:20
• @TerezaTizkova: Yes, this is the standard notation for the disjoint union. Jul 18, 2022 at 19:23
• 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?) Jul 18, 2022 at 23:48

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 $$\beta(-\infty,0]$$.

• No, this argument does not prove the desired claim (which, I think, is actually false). Jul 15, 2022 at 13:01
• @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. Jul 15, 2022 at 13:10
• @MoisheKohan Ah, thank you, you are right. So I am still stuck with this. Jul 16, 2022 at 13:24
• 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 ${\mathbb R}_{\pm}$. Jul 16, 2022 at 15:31
• I will post a proof when I have time... Jul 16, 2022 at 16:50