$$R = \{x : x \in \mathbb{Z} \text{ and either } x \leq −2 \text{ or } x \geq 5\}$$
$$S = \{−3, −2, 4, 5, 6\}$$
$$T = \{x : x \in \mathbb{Z} \text{ and } x \geq 2\}$$
Find $P(R) \cap \{\{−6, −3, 5\}, \, \{4, 5\}, \, \{7\}, \, \{\}, \{−3, 1, 7\}\}$
I think the answer is $\{7\}$, as everything else within the list on the right hand side doesn't fall within the parameters of $R$. Also, if the question was not asking for the power set of $R$ would the answer be $\{\}$?
Find $|P(P(S \cap T))|$
$(S \cap T) = \{4, 5, 6\}$ Since there are $3$ elements in the list, I am guessing that the number of elements within the list is $2^3 = 8$. Would I be correct in assuming so?
For question 1.
The set $R$ is all integers except $-1, 0, 1, 2, 3 \text{ and } 4$, so $P(R)$ is a set of all sets of integers exept containing any of those six numbers. Hence the intersection does not contain $\{4,5\}$ (because $4$ is excluded) and $\{−3,1,7\}$ (because $1$ is excluded). All remaining sets 'survive', so the answer is $$\{\{−6,−3,5\}, \{7\}, \{\}\}.$$ And the empty set is a member of every power set, so it belongs to the answer, too.
For question 2.
The set $S\cap T$ has three elements, so its power set: $P(S\cap T)$ has $2^3=8$ elements
...and the power set of that power set: $P(P(S\cap T))$ has $2^8=256$ elements.