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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.

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$\newcommand{\cl}{\operatorname{cl}}$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.)

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I'm not familiar with the notation $[0,\rightarrow)$. What does this mean? –  user108903 Jan 18 '13 at 21:21
    
@user108903: Same as $[0,\infty)$. –  Brian M. Scott Jan 18 '13 at 21:23
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I finally found something:

For convenience, let $X= \mathbb{R} \backslash (-1,1)$.

First, we show that $(\beta \mathbb{R})\backslash (-1,1)= \beta (\mathbb{R} \backslash (-1,1))$. Let $f_0 : X \to [0,1]$ be a continuous function. Obviously, $f_0$ can be extend to a continuous function $f : \mathbb{R} \to [0,1]$; then we extend $f$ to a continuous function $\tilde{f} : \beta \mathbb{R} \to [0,1]$ and we get by restriction a continuous function $\tilde{f}_0 : (\beta \mathbb{R}) \backslash (-1,1) \to [0,1]$. By the universal property of Stone-Čech compactification, we deduce that $\beta X= (\beta \mathbb{R}) \backslash (-1,1)$.

With the same argument, we show that for an closed interval $I \subset X$, $\beta I = \text{cl}_{\beta X}(I)$.

Let $F_1= \text{cl}_{\beta X} [1,+ \infty)$ and $F_2= \text{cl}_{\beta X} (-\infty,-1]$. Then, $F_1 \cup F_2 = \beta X$, $F_1 \cap F_1= \emptyset$ and $F_1$, $F_2$ are connected.

Indeed, there exists a continuous function $h : X \to \mathbb{R}$ such that $h_{|[1,+ \infty)} \equiv 0$ and $h_{|(-\infty,1]} \equiv 1$, so when we extend $h$ on $\beta X$ we get a continuous function $\tilde{h}$ such that $\tilde{h} \equiv 0$ on $F_1$ and $\tilde{h} \equiv 1$ on $F_2$; therefore $F_1 \cap F_2= \emptyset$.

Finally, $F_1$ and $F_2$ are connected as closures of connected sets.

So $(\beta \mathbb{R} ) \backslash (-1,1)$ has exactly two connected components; in fact, it is the same thing for $(\beta \mathbb{R}) \backslash (-n,n)$. Because $(\beta \mathbb{R}) \backslash \mathbb{R}= \bigcap\limits_{ n\geq 1 } (\beta \mathbb{R}) \backslash (-n,n)$, we deduce that $(\beta \mathbb{R}) \backslash \mathbb{R}$ has exactly two connected components (the intersection of a non-increasing sequence of connected compact sets is connected).

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It looks okay now. Note that you can get connectedness of $F_1$ and $F_2$ even more easily just by observing that the closure of a connected set is connected. –  Brian M. Scott Jan 18 '13 at 22:02
    
@BrianM.Scott: Indeed, thank you! –  Seirios Jan 18 '13 at 22:05
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