Method 1:(kind of Brute Force)👍
$$S = \{1, \{2\}, \{3, 4\}\}$$
The elements belonging to $S$ are:
$1 \in S$
$\{2\} \in S$
$\{3, 4\} \in S$
if cardinality of S is $n$
cardinality of power set is $2^n$
since 8 element we can easily find $P(s)$ it self as
P(S) = { ∅, {1}, {{2}}, {{3, 4}}, {1, {2}}, {1, {3, 4}}, {{2}, {3, 4}}, {1, {2}, {3, 4}} }
C is false
Method 2:(Preferred)👌
By definition, the power set $\mathcal{P}(S)$ is the set of all subsets of $S$. Therefore, if $A \subseteq S$, then $A \in \mathcal{P}(S)$.
(A) $\{1\} \in \mathcal{P}(S)$: Since $1 \in S$, the singleton set $\{1\}$ is a subset of $S$ ($\{1\} \subseteq S$). Thus, $\{1\} \in \mathcal{P}(S)$. (True)
(B) $\{1, \{3, 4\}\} \in \mathcal{P}(S)$: Since $1 \in S$ and $\{3, 4\} \in S$, the set containing both elements is a subset of $S$ ($\{1, \{3, 4\}\} \subseteq S$). Thus, it belongs to $\mathcal{P}(S)$. (True)
(C) $\{\{1, 2\}, \{3, 4\}\} \in \mathcal{P}(S)$: For this set to be in $\mathcal{P}(S)$, its elements must be elements of $S$. The elements here are $\{1, 2\}$ and $\{3, 4\}$. While $\{3, 4\} \in S$, the element $\{1, 2\} \notin S$ (the number $1$ and the set $\{2\}$ are separate elements in $S$, they are not grouped together as $\{1, 2\}$). Therefore, $\{\{1, 2\}, \{3, 4\}\} \not\subseteq S$, which means it cannot be in $\mathcal{P}(S)$. (False)
(D) $\{\{2\}, \{3, 4\}\} \in \mathcal{P}(S)$: Since $\{2\} \in S$ and $\{3, 4\} \in S$, the set containing them is a subset of $S$ ($\{\{2\}, \{3, 4\}\} \subseteq S$). Thus, it belongss to $\mathcal{P}(S)$. (True)
