Are $\beta \mathbb{Q}$ and $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ homeomorphic?

Question

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 $\beta\mathbb{Q}\setminus \mathbb{Q}$.

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.