TITLE: Connectedness of $\beta \mathbb{R}$
QUESTION [6 upvotes]: I am not very familiar with Stone-Čech compactification, but I would like to understand why the remainder $\beta\mathbb{R}\backslash\mathbb{R}$ has exactly two connected components.

REPLY [8 votes]: $\newcommand{\cl}{\operatorname{cl}}$For any Tikhonov space $X$ let $X^*=\beta X\setminus X$. Let $\Bbb H=[0,\to)$ and $\Bbb L=(\leftarrow,0]$. $\Bbb H$ is $C^*$-embedded in $\Bbb R$, so $\cl_{\beta\Bbb R}\Bbb H=\beta\Bbb H$, and of course $\cl_{\beta\Bbb R}\Bbb L=\beta\Bbb L$. Note that if $x\in\Bbb H^*$, and $V$ is a nbhd of $x$ in $\Bbb R$, then $V\cap\Bbb H$ is unbounded. Thus, the Čech-Stone extension $\beta f$ of the function
$$f:\Bbb R\to\left[-\frac{\pi}2,\frac{\pi}2\right]:x\mapsto\tan^{-1}x$$
must map $\beta\Bbb H^*$ to $\pi/2$. Similarly, $\beta f$ must map $\beta\Bbb L^*$ to $-\pi/2$. But $\Bbb R^*=\Bbb H^*\cup\Bbb L^*$, so $\beta f$ maps $\Bbb R^*$ onto a discrete two-point space, and $\Bbb R^*$ cannot be connected.
To complete the argument, we need only show that $\Bbb H^*$ is connected, since clearly $\Bbb L^*\cong\Bbb H^*$. Suppose, to get a contradiction, that there is a continuous surjection $f:\Bbb H^*\to\{0,1\}$. $\Bbb H$ is locally compact, so it is open in $\beta\Bbb H$, and $\Bbb H^*$ is therefore a compact subset of $\beta\Bbb H$. It follows that $\Bbb H^*$ is $C$-embedded in $\beta\Bbb H$ and hence that $f$ extends to a continuous $\hat f:\beta\Bbb H\to\Bbb R$. Fix $p,q\in\Bbb H^*$ such that $f(p)=0$ and $f(q)=1$, and let $$U=\hat f^{-1}\left[\left(-\frac14,\frac14\right)\right]\quad\text{and}\quad V=\hat f^{-1}\left[\left(\frac34,\frac54\right)\right]\;.$$
$U\cap\Bbb H$ and $V\cap\Bbb H$ are both unbounded, so the intermediate value theorem ensures that $$A=\left\{x\in\Bbb H:\hat f(x)=\frac12\right\}$$ is unbounded. But then $f(x)=\hat f(x)=\frac12$ for some $x\in\Bbb H^*$, which is the desired contradiction.
It follows that $\Bbb H^*$ and $\Bbb L^*$ are the two components of $\Bbb R^*$.
(This is adapted from Gillman & Jerison, Rings of Continuous Functions, which is still very useful, despite being over $50$ years old now.)