Given $A$ is a set of cardinality $c$ and $p, q$ and $r$ are distinct members of $A$, show that there are disjoint subsets $P, Q$ and $R$ of $A$ each with cardinality $c$ such that $p \in P,q \in Q$ and $r \in R$.
Even though this statement seems intuitive, could someone provide a complete, formal proof for it?
I can show that an infinite set always has a countably infinite subset by removing its elements one-by-one and then demonstrating that the union of these elements is a subset of the original infinite set in a nutshell. But how do I show that each of these distinct and disjoint subsets have cardinality c and that each of $p, q$ and $r$ are the members of these subsets respectively? Even if someone could provide an alternative proof that would be helpful.
As #A = c, there is a bijection f:A -> R.
Wlog assume $f(p) < f(q) < f(r)$,
Find a,b with $f(p) < a < f(q) < b < f(r)$,
Since a and b divide R into three sets with cardinality c,
each containing f(p), f(q), f(r), respectively,
the inverse by f of those sets provides the sets you want.
If A had an infinite cardinality of k with distinct p,q,r in A,
is there three pairwise disjoint sets P, Q, R of cardinality k
with p in P, q in Q, r in R?
The key for that is using k + k + k = k to get a bijection of A into three pairwise disjoint subsets of A of cardinality k.