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Let $X$ be an uncountable set endowed with the discrete topology. Let $\mathcal{P}(X)$ be the set of all Borel probability measures on $X$, and consider the subset $A$ of $\mathcal{P}(X)$ consisting of all the probability measures with finite support. Can I say something about $A$? Is it compact (if I endow $\mathcal{P}(X)$ with the weak* topology)? The only thing that I could say, if my reasoning is correct, is that $A$ is open.

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    $\begingroup$ What do you mean by the weak$^*$ topology? $\endgroup$ – Jochen Nov 29 '15 at 14:50
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Endowing $X$ with the discrete topology is in a sense a "degenerate" case, because all functions will be continuous and the delta measure will be a Borel measure.

Even in the case that $X=[0,1]$, and $\delta_n(A)=\begin{cases}1& 1/n\in A\\0&\mathrm{ \ otherwise}\end{cases}$, any bounded $f$, in particular one for which $\lim_{x\rightarrow0^+}f(x)$ exists but is not equal to $f(0)$, will still be continuous since we have the discrete topology on $[0,1]$. Thus for this sequence of measures, the only possible limit is $\delta$, with support at $0$, and so $$\lim_{n\rightarrow\infty}\int f d\delta_n=\lim_{n\rightarrow\infty}f(1/n)\neq f(0)=\int fd\delta.$$ Thus this sequence measures does not even converge in the weak sense. That is, $\mathcal{P}(X)$ is not even closed in the weak* topology, so we can't hope for much structure in this space...

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