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Given a compact Hausdorff space $X$, does there exist a probability $\mu$ on X such that the support of $\mu$ is $X$? This is equivalent to say, for any unital commutative C*-algebra, can we show the existence of a faithful state?

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Just an observation: If the space is separable, you can choose a countable, dense sequence $\{x_n\}$ and define $\mu$ by $\mu(E)=\sum 2^{-n}\chi_E(x_n)$. I don't know about the general case though. –  Nick Strehlke Oct 29 '11 at 16:35
    
$X$ may be not separable, for example if $X=[0,1]^{[0,1]}$ endowed with the topology of uniform convergence. How is the support defined? –  Davide Giraudo Apr 22 '12 at 16:06
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@DavideGiraudo: Note that your $X$ is not compact. –  Nate Eldredge Apr 22 '12 at 16:47
    
Sorry, I meant pointwise convergence. –  Davide Giraudo Apr 22 '12 at 16:50
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@DavideGiraudo: Note that your $X$ is separable :) –  t.b. Apr 22 '12 at 16:59

1 Answer 1

In general, the answer is no.

It is maybe easiest to see for your second question: let $\Omega$ be an uncountable set and consider the $C^*$-algebra $B(\Omega)$ consisting of all bounded complex-valued functions on $\Omega$ (with the supremum norm). If $\ell$ were a faithful state, we would have to have $\ell(1) \ge \sum_{\omega \in \Omega} \ell(1_{\{\omega\}})$ which is an uncountable sum of positive terms and cannot be finite. (You can also think of $B(\Omega)$ as $L^\infty(\Omega)$ with counting measure.)

Equivalently, let $X = \beta \Omega$ be the Stone-Cech compactification of $\Omega$, where we give $\Omega$ the discrete topology. (So $C(X) = B(\Omega)$.) Then for each $\omega \in \Omega$, $\{\omega\}$ is open in $X$. So any measure on $X$ that charges every open set must have total mass infinity. It cannot even be $\sigma$-finite.

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