Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
I'm a bit confused here, but isn't $X=\beta X$? –  Gina Jan 3 '14 at 12:12
$\beta X$, as a set, constists of all tupels $(f(x))_{f \in C(X,[0,1])}$ where $x \in X$. –  Andre Jan 3 '14 at 12:15
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 '14 at 12:27
Oh yes, sorry. You are absolutely right. –  Andre Jan 3 '14 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 '14 at 20:00

1 Answer 1

up vote 3 down vote accepted

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) $\beta X$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.