Question on compactness of $\Delta(X)$ with the weak$^{\ast}$ topology for compact $X$ Let $X$ be a metric space. $\Delta(X)$ is the set of all probability Borel measures on $X$(which is regular, of course).$\Delta(X)$ is endowed with weak$^{\ast}$ topology for which the mapping $\mu \mapsto \mu(f)= \int f d\mu$ is upper-semicontinuous for all bounded upper-semicontinuous $f$.
An exercise in Jean-François Mertens, Sylvain Sorin and Shmuel Zamir(2000)(page 13) is to show that if $X$ is compact, then $\Delta(X)$ is compact.
I know that "if" in this assertion can be replaced by iff. That's because that  $\mu \mapsto \mu(f)= \int f d\mu$ is u.s.c for all bounded u.s.c $f$ is equivalent to that $\mu \mapsto \mu(f)= \int f d\mu$ is continuous for all bounded continuous $f$ as shown in this note (proposition 1.12, page 6). $\Delta(X)$ equipped with the latter topology is compact, iff $X$ is compact (Theorem 6.4, page 45, K.R. Parthasarathy(1967) see here).
But I can't understand the hint on page 13 in the book, especially the first step.

Question: Isn't it the case that proposition $a$ which is asked to be proved followed from the definition of $M(X)$? If there is no $\mu$ for some $p$, then $p$ must be in $M(X)$, right? Moreover, I can't see why $M(X)$ exists.
I also checked Mertens(1986) (paywall), in which the proof is identical to the hint here.
 A: $M(X)$ exists since it is simply the set of elements of $P$ that are minimal. It could be empty though, but it follows from a) that it is not. Proposition a) says more than the definition. If you take $\mathbb{Z}$ with the usal ordering, there is no element $p\in\mathbb{Z}$ such that some minimal element of $\mathbb{Z}$ is smaller- because the set of minimal elements is empty. Similarly, the set of minimal elements smaller than some $p$ may be empty in the problem, and you have to show that this is not the case to prove a).
For the problem at hand, take some $p\in P$. Let $Q=\{q\in P:q\leq p\}$. We want to show that $Q$ has a minimal element. This follows from Zorn's lemma if we can show that every chain in $Q$ is bounded below. By the hint, we know that the values of functions in the chain are bounded below. So the pointwise infimum of the chain exists. To finish the argument, you have to show that this pointwise infimum belongs again to $P$.
Remark: Learning from that book is pure masochism. It is mainly a very advanced reference work.
