Compact subset of infinite-dimensional space has empty interior My question is related to this question.
My space is the set of all Borel probability measures on $\Theta=[0,1]$, which is a compact metric space under the Prokhorov metric. Call this space $\Delta \Theta$. This is a compact subset of the space of all signed measures on $[0,1]$ which is a vector space with the total variation norm. As we know, the interior of a compact set in an infinite dimensional TVS is empty. So that means $int(\Delta \Theta)=\emptyset$. So $\Delta \Theta$ cannot contain any open sets. Take any continuous function $f: \Delta \Theta \rightarrow \mathbb{R}$. The inverse image of any open set in $\mathbb{R}$ under $f$ - which is, by definition, a subset of $\Delta \Theta$ - is an open set by continuity of $f$. This seems like a contradiction and I'm totally confused.
I think the apparent contradiction is arising because of the change of topology - the Prokhorov metric on the $\Delta \Theta$ space is not the same as the metric induced by the TV norm on the VS in which $\Delta \Theta$ lives. And maybe under the TV metric $\Delta \Theta$ is not compact?
Any help in resolving this is highly appreciated. Thanks in advance.
 A: Let $M\Theta$ denote the vector space of all signed measures on $\Theta$.
There are (at least) two possible topologies on $M \Theta$.  One is the total variation topology TV.  Indeed $\Delta \Theta$ is not compact in this topology; for instance, the set of unit point masses are all at total variation distance 2 from each other, so they form an infinite set with no limit points.  In particular, the topology induced on $\Delta \Theta$ by the Prokhorov metric does not coincide with the subspace topology induced from TV.  It also has empty interior, since every open set of $M \Theta$ with respect to the TV topology must contain measures with distinct total variations.
(In a previous version I said $\Delta \Theta$ had nonempty interior; but I was mixing it up with the closed unit ball of $M\Theta$ with respect to TV, which would be the set of all signed measures having total variation $\le 1$.)
Another topology on $M \Theta$ is the weak-* topology W*, induced by its being the dual of the space $C(\Theta)$ of continuous functions.  Now $\Delta \Theta$ is a compact subset of $M\Theta$ with respect to the W* topology, and the Prokhorov metric does coincide with the subspace topology from W*.  However $\Delta \Theta$ again has empty interior with respect to the topology W* of $M \Theta$.
What your argument shows is that $\Delta \Theta$ has nonempty interior in itself, with respect to the subspace topology induced on it by W*.  That is of course trivial; every topological space is open with respect to its own topology.  But it isn't an open subset of $(M\Theta, W^*)$, and moreover it doesn't contain any such open subsets.
For your argument involving $f$ to work, you'd have to start with a function $f$ whose domain is not only $\Delta \Theta$ but actually $M \Theta$, and which is continuous with respect to the W* topology on $M \Theta$.  And then it will fail.  The claim is that the preimage of any open subset of $\mathbb{R}$ will either be empty, or will contain points of $M \Theta$ that are outside $\Delta \Theta$.
Actually, it is not too hard to show that any nonempty open subset $U$ of $(M\Theta, W^*)$ contains measures of arbitrarily large total variation, and so cannot be a subset of $\Delta \Theta$.  In particular, there will always exist $\mu \in U$ such that $c\mu \in U$ for every $c \in \mathbb{R}$, i.e. $U$ contains the "line" $\mathbb{R} \mu$.  Indeed, it follows from the definition of the weak-* topology that there exist finitely many $f_1, \dots, f_n \in C(\Theta)$ and $\epsilon > 0$ such that if $\left|\int f_i \,d\mu\right| < \epsilon$ for all $i=1,\dots, n$, then $\mu \in U$.  But with a little linear algebra, you can find a nonzero measure $\mu$ such that $\int f_i \,d\mu = 0$ for all $i=1,\dots, n$.  Then $c \mu$ has the same property and is therefore in $U$.
