Let $A = \{1\}$, $B = \{1,\{1\}\}$, $C = \mathcal{P}(A)$, and $D = \mathcal{P}(B).$ Which one is true? Which one is true?


*

*$C \in D.$

*$C \subseteq D.$

I can't quite wrap my head around so many curves, is a power set an element? It should be a subset I believe but can a power set be an element?
So $C = \{\emptyset, 1\}$ and $D = \{\emptyset, 1, \{1\}, \{1, \{1\}\}\}.$
Does this mean that both $1$ and $2$ are false because ${\emptyset, 1\}$ doesn't exist in $D?$ Can someone please confirm, thank you!
 A: With $A=\{1\}$ and with $B=\{1,\{1\}\}$ we have:
$C = \mathcal{P}(A) = \{\emptyset,\{1\}\}$
$D = \mathcal{P}(B) = \{\emptyset,\{1\},\{\{1\}\},\{1,\{1\}\}\}$
To make matters clearer, let us use colors for each of the elements of $D$.  Now, we can see that $C=\color{blue}{\{\emptyset,\{1\}\}}\notin \{\color{orange}{\emptyset},\color{red}{\{1\}},\color{green}{\{\{1\}\}},\color{purple}{\{1,\{1\}\}}\}=D$.  Since $C$ is not equal to the orange element, the red element, green element, or the purple element we have that $C$ is not equal to any of the elements of $D$ and so $C$ is not considered an element of $D$.  Note, the entire object I colored in purple here is collectively considered to be one of the elements of $D$, this despite the fact that it contains smaller pieces inside of it.  Note also how the green element is different than the red element, that the positioning of the braces and "depth" is relevant and makes these distinguishable from one another.
On the other hand, $C=\{\color{orange}{\emptyset},\color{red}{\{1\}}\}\subseteq \{\color{orange}{\emptyset},\color{red}{\{1\}},\color{green}{\{\{1\}\}},\color{purple}{\{1,\{1\}\}}\}=D$ since each element in $C$ is also an element in $D$.

Now... while $C\notin D$ but $C\subseteq D$ happened to be true, it is possible to have another set such as $E=\{\emptyset, \{1\},\{\emptyset,\{1\}\}$ which does have $C$ both as an element and as a subset simultaneously.
A: Let $A = \{1\}$, $B = \{1,\{1\}\}$, $C = \mathcal{P}(A)$, and $D = \mathcal{P}(B).$
Then we have that $C = \mathcal{P}(A) = \{\emptyset, \{1\}\}$ and $D = \mathcal{P}(B) = \{\emptyset, \{1\},\{\{1\}\},\{1,\{1\}\}\}.$
Since $\emptyset$ and $\{1\}$ (elements of $C$) are both elements of $D,$ we conclude that $C \subseteq D.$

Also, note that: If $A$ and $B$ are sets such that $A \subseteq B,$ then $\mathcal{P}(A) \subseteq \mathcal{P}(B).$
Proof: Suppose $A \subseteq B.$ Let $X \in \mathcal{P}(A).$ Then $X \subseteq A.$ From our hypothesis follows that $X \subseteq B,$ and by definition, $X \in \mathcal{P}(B).$ $\square$

To have that $C \in D,$ we would need to have the $\{\emptyset, \{1\}\}$ as an element of $D,$ which is not the case.
