The canonical inclusion $\beta\mathbb{Q}\setminus \mathbb{Q} \hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of $\beta\mathbb{Q}\setminus \mathbb{Q}$. Even so, this doesn't necessarily mean that $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ and $\beta\mathbb{Q}$ are not homeomorphic, just that this particular map doesn't work for this compactification, so that they might be homeomorphic less "canonically". For example, $\beta(\mathbb{Q}\setminus\{0\})\cong \beta\mathbb{Q}$ but the map $\mathbb{Q}\setminus\{0\}\hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of $\mathbb{Q}\setminus\{0\}$, even though it's a dense embedding into a compact space homeomorphic to $\beta(\mathbb{Q}\setminus\{0\})$.

I've tried looking online for some properties of $\beta\mathbb{Q}\setminus\mathbb{Q}$ that would exclude a homeomorphism, though while looking I could only find properties it has in common with $\mathbb{Q}$, with exception of homogeneity (though I don't think this particular property amounts to much). For example, $\beta\mathbb{Q}\setminus \mathbb{Q}$ is zero-dimensional, not extremally disconnected.

As in the title, is $\beta\mathbb{Q}\cong \beta(\beta\mathbb{Q}\setminus \mathbb{Q})$?

Edit: By suggestion of @R. van Dobben de Bruyn's comment, I've checked cardinalities of both spaces. In the article "The Stone-Čech compactification of the rational world" by M. P. Stannett it's shown that $\beta\mathbb{Q}\setminus\mathbb{Q}$ is separable, so that $$\lvert\beta(\beta\mathbb{Q}\setminus\mathbb{Q})\rvert \leq 2^{2^{d(\beta(\beta\mathbb{Q}\setminus\mathbb{Q}))}} \leq 2^{2^{d(\beta\mathbb{Q}\setminus\mathbb{Q})}} = 2^\mathfrak{c}$$ while the inclusion $\beta\mathbb{Q}\setminus \mathbb{Q}\hookrightarrow \beta\mathbb{Q}$ induces a surjection $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})\to \beta\mathbb{Q}$, thus $$\lvert\beta\mathbb{Q}\rvert = \lvert\beta(\beta\mathbb{Q}\setminus\mathbb{Q})\rvert = 2^\mathfrak{c}.$$ This shows in particular that there's no issue with cardinality. I think trying to approach it with weight would provide similar results, though I haven't checked that.

Edit2: Here's my proof of my claim that $\beta\mathbb{Q}\setminus\mathbb{Q}\hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of the space $\beta\mathbb{Q}\setminus\mathbb{Q}$. Note that $\beta\mathbb{Q}\setminus\mathbb{Q}$ is dense in $\beta\mathbb{Q}$ since $\mathbb{Q}$ is nowhere locally compact. The space $\beta\mathbb{Q}\setminus \mathbb{Q}$ is not $C^*$-embedded in $\beta\mathbb{Q}$: the decomposition $$\beta\mathbb{Q}\setminus \{0\} = (\overline{\mathbb{Q}}_+\setminus\{0\}) \cup (\overline{\mathbb{Q}}_-\setminus \{0\})$$ of $\beta\mathbb{Q}\setminus\{0\}$ where $\mathbb{Q}_+ = (0, \infty)\cap\mathbb{Q}$ and $\mathbb{Q}_- = (-\infty, 0)\cap \mathbb{Q}$ into two disjoint closed sets in $\beta\mathbb{Q}\setminus\{0\}$ shows that $\DeclareMathOperator\sgn{sgn}\sgn:\mathbb{Q}\setminus\{0\}\to \mathbb{R}$ can be continuously extended to $\sgn^*:\beta\mathbb{Q}\setminus\{0\}\to\mathbb{R}$ but clearly not to whole of $\beta\mathbb{Q}$. If the function $\sgn^*\restriction_{\beta\mathbb{Q}\setminus\mathbb{Q}}$ were to continuously extend to $\beta\mathbb{Q}$, the extension would have to be equal to $\sgn^*$ on $\beta\mathbb{Q}\setminus\{0\}$, which is impossible since $\sgn^*$ doesn't extend to $\beta\mathbb{Q}$. Since for $A\subseteq \beta\mathbb{Q}$, we have that $\DeclareMathOperator\cl{cl}\cl_{\beta\mathbb{Q}}A =\beta A$ (that is $A\hookrightarrow \cl_{\beta\mathbb{Q}} A$ is the Stone-Čech compactification of $A$) iff $A$ is $C^*$-embedded in $\beta\mathbb{Q}$, the inclusion $\beta\mathbb{Q}\setminus \mathbb{Q} \hookrightarrow \beta\mathbb{Q}$ isn't the Stone-Cech compactification of $\beta\mathbb{Q}\setminus \mathbb{Q}$.

  • 2
    $\begingroup$ At a first glance, it seems to me they might have wildly different cardinalities. Or did you try this already? $\endgroup$ May 4, 2023 at 19:20
  • $\begingroup$ Can you justify or provide a reference for the statement of the first sentence? Is it supposed to be a well-known or trivial fact? $\endgroup$
    – Gro-Tsen
    May 4, 2023 at 21:18
  • 1
    $\begingroup$ @Gro-Tsen Yes, see edit2 $\endgroup$
    – Jakobian
    May 4, 2023 at 21:35
  • 2
    $\begingroup$ For each partition of $\beta\mathbb{Q}$ into two components $A,B$, one of those components has a non-trivial convergent sequence. I don't see the same thing happening with $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ but I am currently too busy to write and check the proof. $\endgroup$ May 5, 2023 at 0:18
  • 2
    $\begingroup$ I don't know where this is explicitly stated--it might be in the Walker book on the Stone-Cech compactification--but it is not hard. If $p$ is a point of first countability of $\beta X \setminus X$, then $\beta X \setminus \{p\}$ is $\sigma$-compact and therefore normal. But if $a = (a_n)$ a sequence in $X$ that converges to $p$, then $\{a_n: n = 1, 2, ...\}$ is closed in $X$ but a function which alternates between $0$ and $1$ on that set does not extend to $p$. $\endgroup$
    – Anonymous
    May 5, 2023 at 13:42

1 Answer 1


Let me summarize the discussion in the comments as an answer. Let $\chi(x, Y)$ be the character of $x$ in $Y$ i.e. the least cardinality of a local basis of the point $x$ in space $Y$.

Proposition 1. If $p\in \beta X\setminus X$ then $\chi(p, \beta X)$ is uncountable.

Proof: If it werre $\chi(p, \beta X) = \aleph_0$ we would find a sequence $(a_n)\subseteq X$ with $a_n\to p$ and $a_n\neq a_m$ for $n\neq m$. Since $\beta X\setminus \{p\}$ is $\sigma$-compact, it's Lindelöf regular, so normal. Moreover, $A = \{a_n : n\in\mathbb{N}\}$ is a closed discrete subset of $\beta X\setminus \{p\}$. Thus the function $f:A\to [-1, 1]$ given by $f(a_n) = (-1)^n$ has a continuous extension $\tilde f$ to $\beta X\setminus \{p\}$, but no extension to $\beta X$. This is a contradiction, because $\tilde f\restriction_X$ needs to continuously extend to $\beta X$, and thus the extension needs to be equal to $\tilde f$ on $\beta X\setminus \{p\}$.

Proposition 2. If $S\subseteq X$ is dense, $X$ regular, $p\in S$, then $\chi(p, S) = \chi(p, X)$.

This can be found in the Handbook of Set-theoretic Topology.

Proposition 3. If $p\in \beta(\beta\mathbb{Q}\setminus \mathbb{Q})$ then $\chi(p, \beta(\beta\mathbb{Q}\setminus \mathbb{Q}))$ is uncountable.

Proof: If $p\in \beta(\beta\mathbb{Q}\setminus\mathbb{Q})\setminus (\beta\mathbb{Q}\setminus\mathbb{Q})$ then $\chi(p, \beta(\beta\mathbb{Q}\setminus\mathbb{Q}))$ is uncountable by proposition 1. If $p\in \beta\mathbb{Q}\setminus\mathbb{Q}$, then $\chi(p, \beta(\beta\mathbb{Q}\setminus\mathbb{Q})) = \chi(p, \beta\mathbb{Q}\setminus\mathbb{Q}) = \chi(p, \beta\mathbb{Q})$ by proposition 2, which is uncountable, again by proposition 1.

Thus $\beta\mathbb{Q}$ has elements of countable character while $\beta(\beta\mathbb{Q}\setminus \mathbb{Q})$ has no such points, so the two spaces are not homeomorphic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.