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Let $X$ be a discrete space and $\beta X$ its Stone-Čech-compactification, given by $\overline{\iota(X)}$ where $$ \iota:\ X \to \prod_{f \in C(X,[0,1])} [0,1],\quad x \mapsto (f(x))_{f \in C(X,[0,1])}. $$ I have already proven that the linear span of idempotent functions in $C_b(X,\mathbb R)$ lies dense in $C_b(X,\mathbb R)$ (in sup-norm). Since $X$ is discrete these are only the bounded functions. How can I use this fact to prove that $\beta X$ is totally disconnected, i.e. the connected components consist of only one point ?

It would also be sufficient to prove that $\beta X$ is totally separated, i.e. if $x \neq y$ in $\beta X$ then there is a clopen $C \subset \beta X$ s.t. $y \notin C \ni x$.

I would appreciacte some hints, in particular why we need idempotent functions.

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I'm a bit confused here, but isn't $X=\beta X$? –  Gina Jan 3 at 12:12
    
$\beta X$, as a set, constists of all tupels $(f(x))_{f \in C(X,[0,1])}$ where $x \in X$. –  André Jan 3 at 12:15
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I think you meant to say that $\beta X$ is the closure of that set. Because given any point $x$ in $X$ there exist a function that give $1$ to that point and $0$ to the rest (continuous since $X$ is discrete), which means that there exist an open set containing only $i(x)$ and no other point in $i(X)$. So if you have $\beta X$ being just merely $i(X)$ rather than the closure, then $\beta X$ is discrete and have the same cardinality as $X$, ie. $X=\beta X$. –  Gina Jan 3 at 12:27
    
Oh yes, sorry. You are absolutely right. –  André Jan 3 at 13:07
    
Um, can you explain what idempotent function is? It isn't the usual idempotent isn't it (ie. self-composition is the same as itself)? –  Gina Jan 3 at 20:00

1 Answer 1

First of all, identify $X$ with its image $\iota(X)$ in $\beta X$. Now, remember the universal property for $\beta X$:

For every $f\in C_b(X,\mathbb{R})$, there exists a unique $\overline{f}\in C(\beta X,\mathbb{R})$ extending $f$.

Notice that, if $f\in C_b(X,\mathbb{R})$ is idempotent, then so is its extension $\overline{f}\in C(\beta X,\mathbb{R})$ (to prove this, use the facts that $\mathbb{R}$ is Hausdorff and $X$ is dense in $\beta X$). Also, since $X$ is dense in $\beta X$, then, for any $F,G\in C(\beta X,\mathbb{R})$, we have $$\sup\left\{|F(x)-G(x)|:x\in\beta X\right\}=\sup\left\{|F(x)-G(x)|:x\in X\right\}.$$

With that, we conclude that the linear span of the idempotents is dense in $C(\beta X,\mathbb{R})$ with sup-norm.

Now, let $x\neq y$ in $\beta X$. By the above assertion (and Urysohn's Lemma), there exists an idempotent $g\in C(\beta X,\mathbb{R})$ such that $g(x)\neq g(y)$. It follows easily that, for any subset $Y\subseteq \beta X$ containing $x$ and $y$, we have $g(Y)=\left\{0,1\right\}$, which is disconnected. Since $g$ is continuous, $Y$ cannot be connected.

Concluding, any subset of $\beta X$ with 2 or more elements cannot be connected. This means that $\beta X$ is totally disconnected.


This is a particular case of a more general property: If $A$ is any Banach algebra in which the linear span of the idempotents is dense, then its spectrum (if not empty) is totally disconnected (if we change $\mathbb{R}$ by $\mathbb{C}$, but this is not a problem...). In the problem you gave, $A=C_b(X,\mathbb{R})\sim C(\beta X,\mathbb{R})$ and its spectrum is (homeomorphic to) $\beta X$.

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