Given a metrizable space Z, I know that the set of probability measures on Z, call it $P(Z)$, upon which we impose the weak-* topology, is separable since by the Krein-Milman Theorem, the set of simple probability measures is dense in $P(Z)$.
Is it the case that $P(Z)$ is also connected? What conditions would have to be put on $Z$ to make this true?
For example, if $Z \subseteq \mathbb{R}$, then since $\mathbb{R}$ is connected, it seems that $P(Z)$ should be too. But, I don't know how to prove this.
Take any two probability measures $\mu$ and $\nu$. For every $r\in [0,1]$, the convex combination $r\lambda+(1-r)\nu$ given by $$(r\lambda+(1-r)\nu)(A)=r\lambda(A)+(1-r)\nu(A)$$ for every Borel set $A$ is again a probability measure. The function $r\mapsto r\lambda+(1-r)\nu$ is easily seen to be continuous. It follows that the space of probability measures is path-connected and therefore also connected.
The argument always works if you endow the space of Borel probability measures on your metric space with the weakest topology such that the function $\mu\mapsto \int g~\mathrm d \mu$ is continuous for every bounded continuous function $g$ from your metric space to the reals. But for general metric spaces, this doesn't give you something you can identify with a subset of a dual space endowed with the weak*-topology. If your metric space is not separable, this topology is not going to be separable either.