Proof the proposition.
For this question. I have no idea how to start.
The P(I) just messes me up.
Since J belongs in the set P(I), shouldn't the statement be ⊆ instead of the top one?
which is all K ∈ I, X∈ Sk
which is all K ∈ J, X∈ Sk
Some hints to get me start the question would be helpful. Thank you.
Imagine that $I$ and $J$ are groups of people and $J\subseteq I$. Also, for $k\in I$, let $S_k$ be the set of days where person $k$ is free to come to a movie night. To find the set of days that work for all people in $I$, we take $\bigcap_{k\in I}S_k$, and similarly for $J$. Now, clearly, the more people, the harder is to find a night that suits everyone, i.e., the smaller the intersection of $S_k$ is. Formally,
$$J\subseteq I\implies \bigcap_{k\in I}S_k\subseteq\bigcap_{k\in J}S_k.$$