Proper refinmene of two sets Assuming that P and S are partitions of the same set with a finite number of elements, if $P$ I a proper refinement of $S$, both defined on the same set, then $P$ has more blocks/has higher cardinality then $S$?
 A: Let $\mathscr{P}$ and $\mathscr{S}$ be finite partitions of a set $X$. $\mathscr{P}$ is a refinement of $\mathscr{S}$ if every block of $\mathscr{P}$ is contained in a block of $\mathscr{S}$. Thus, if $\mathscr{P}$ is a refinement of $\mathscr{S}$, there is a function $f:\mathscr{P}\to\mathscr{S}$ such that $P\subseteq f(P)$ for each $P\in\mathscr{P}$. Since $\bigcup\mathscr{P}=X=\bigcup\mathscr{S}$, $f$ is surjective, and it’s easy to see that every block of $S$ is a union of blocks of $P$: specifically,
$$S=\bigcup\{P\in\mathscr{P}:f(P)=S\}\tag{1}$$
for each $S\in\mathscr{S}$.
$\mathscr{P}$ is a proper refinement of $\mathscr{S}$ if it is a refinement of $\mathscr{S}$ and is not equal to $\mathscr{S}$. If the map $f$ is a bijection, $(1)$ implies that $\mathscr{P}=\mathscr{S}$. Thus, if $\mathscr{P}$ is a proper refinement of $\mathscr{S}$, $f$ cannot be a bijection. Thus, if $\mathscr{P}$ is a proper refinement of $\mathscr{S}$, we have a surjection $f:\mathscr{P}\to\mathscr{S}$ that is not a bijection, and it is a basic fact about finite sets this implies that $|\mathscr{P}|>|\mathscr{S}|$.
You could think about it this way: since $f$ is surjective but not a bijection, it cannot be injective, so there are $P,Q\in\mathscr{P}$ such that $f(P)=f(Q)$, but $P\ne Q$. If we let $S=f(P)$, then $P$ and $Q$ are distinct subsets of $S$, i.e., $S$ is a block of $\mathscr{S}$ that contains at least two blocks of $\mathscr{P}$. Each block of $\mathscr{S}$ contains at least one block of $\mathscr{P}$, and $S$ contains at least one extra block of $\mathscr{P}$, so $\mathscr{P}$ must have at least one more block than $\mathscr{S}$.
