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

3more comments