Stone–Čech compactification of $\mathbb{N}, \mathbb{Q}$ and $\mathbb{R}$ I'm trying to find connections between Stone–Čech compactifications of $\mathbb{N}, \mathbb{Q}$ and $\mathbb{R}$, all with the euclidean topology. So, are there any ? e.g. is $\beta \mathbb{Q} = \beta \mathbb{R}$ ? I know that the cardinality of $\beta \mathbb{N} = \beta \mathbb{Q} =  \beta \mathbb{R}$ thanks to Wikipedia, but how does one prove it ?
 A: In answer to the first part of your question, it's not true that $\beta\mathbb Q=\beta\mathbb R$. For suppose $g$ is some autohomeomorphism of $\mathbb R$. This extends to an autohomeomorphism $\beta g$ of $\beta\mathbb R$, and if we had $\beta\mathbb  Q=\beta\mathbb R$, it would follow that $\beta g$ is an autohomeomorphism of $\beta\mathbb Q$. Since the restriction of $\beta g$ to $\mathbb Q$ is then an autohomeomorphism of $\mathbb Q$ (see the paper cited below), it would follow that every autohomeomorphism of $\mathbb R$ maps $\mathbb Q$ to itself, and this is clearly not true (e.g., $x \mapsto x +\sqrt2$).
Re. comparison of the compactifications, it's potentially worth looking at the remainders. It turns out, for example, that $\mathbb Q$* is separable but $\mathbb R$* isn't. See: M. P. Stannett, "The Stone-Čech Compactification of the Rational World", Glasgow Math. J., vol. 30 (1988) pp. 181-188.
A: Corrected Version: Let $f$ be any map of $\mathbb{N}$ onto $\mathbb{Q}$. Then $f$ is a continuous map of $\mathbb{N}$ into $\beta\mathbb{Q}$, so it extends to a map $\beta f:\beta\mathbb{N}\to\beta\mathbb{Q}$. Since $f$ is a surjection, so is $\beta f$, and it follows that $|\beta\mathbb{N}|\ge|\beta\mathbb{Q}|$. Similarly, the inclusion map $i$ of $\mathbb{Q}$ into $\mathbb{R}$ has a continuous extension to $\beta i:\beta\mathbb{Q}\to\beta\mathbb{R}$, and since $i[\mathbb{Q}]$ is dense in $\mathbb{R}$, $\beta i$ is a surjection, and $|\beta\mathbb{Q}|\ge|\beta\mathbb{R}|$. Finally, $\mathbb{N}$ is $C^*$-embedded in $\mathbb{R}$, so $\beta\mathbb{N}$ embeds in $\beta\mathbb{R}$, and $|\beta\mathbb{N}|\le|\beta\mathbb{R}|$. Thus, $|\beta\mathbb{N}|=|\beta\mathbb{Q}|=|\beta\mathbb{R}|$, and I recently gave a proof that $|\beta\mathbb{N}|=2^\mathfrak{c}$ in this answer.
Note that we knew from the beginning that these cardinalities were at most $2^\mathfrak{c}$: $\mathbb{N}$, $\mathbb{Q}$, and $\mathbb{R}$ are separable and dense in their respective Čech-Stone compactifications, so $\beta\mathbb{N},\beta\mathbb{Q}$, and $\beta\mathbb{R}$ are also separable, and a separable Hausdorff space cannot have more than $2^{2^\omega}=2^\mathfrak{c}$ points.

To see that if $X$ is a separable Hausdorff space, then $|X|\le 2^\mathfrak{c}$, let $D$ be a countable dense subset of $X$. For each $x\in X$ let $\mathscr{D}(x)=\{V\cap D:V\text{ is an open nbhd of }x\}$; $\mathscr{D}(x)$ is a subset of $\wp(D)$, and $|\wp(D)|=\mathfrak{c}$, so there are only $2^\mathfrak{c}$ possible families $\mathscr{D}(x)$. Hausdorffness of $X$ implies that $\mathscr{D}(x)\ne\mathscr{D}(y)$ when $x\ne y$, so $|X|\le 2^\mathfrak{c}$.

