Let $X$ be a compact metric space and denote $\mathcal M(X)$ the set of probability measures on $X$. For $\mu\in\mathcal M(X)$ we write $\operatorname{supp} \mu$ for the *support of $\mu$*. As is well known, if a sequence of measures $(\mu_n)_{n=1}^\infty$ weak$^*$ converges to some $\mu\in\mathcal M(X)$, then
$$
\operatorname{supp}\mu\subset \lim_{n\to\infty}(\operatorname{supp}\mu_n).
$$
and the inverse is not necessarily true.

We can define the following metric on $\mathcal{M}(X)$:

$$ d^{PH}(\mu,\mu_n)=\text{max}\{d^P(\mu,\mu_n),d^H(\mu,\mu_n)\}$$
where $d^P(\mu,\mu_n)$ is the Prokhorov metric on $\mathcal{M}(X)$ and $d^H(\mu,\mu_n)$ is a metric between $\mu$ and $\mu_n$ defined by the Hausfdorff distance between $\operatorname{supp}\mu$ and $\operatorname{supp}\mu_n$.

Finally denote as $d^{TV}(\mu, \mu_n)$ the total variation metric on $\mathcal{M}(X)$, given by 
$$d^{TV}(\mu, \mu_n)= \sup_{A \in \mathcal F}|\mu_n (A) - \mu(A)| $$ where $\mathcal F$ is the underlying $\sigma$-algebra on $X$.

I have the following question: Is it true that the topology on $\mathcal{M}(X)$ generated by $d^{TV}(\cdot)$ is strictly finer than the topology generated by $d^{PH}(\cdot)$ and how could one prove this? Is the topology generated by $d^{PH}(\cdot)$ separable?