If $B \subseteq A$, why is $\{ B \}$ not in the power set of $A$? Define $B = \{1\}$ and $A=\{1, 2\}$.  Then the power set of $A$ is $\mathcal P(A)= \{ \emptyset, \{1\}, \{2\}, A \}$.  Let $C= \{ B \} = \{ \{1 \}\}$.  Why is $C \notin \mathcal P(A)$?  
I dont see why $C$ is not an element of $\mathcal P(A)$, because with the brackets around B you get the set of the set containing B. In the Power Set of A i see that we have a set containing the set with 1. Is it wrong because the one in the power set isnt the set containing the set with 1. I dont know how else to read the power set though. Its a set with the element set containing one.
 A: I like to think of simple sets as boxes.  Then $B$ is a box that contains the number $1$.  And $\{B\}$ is a box that contains a box that contains the number $1$.  So, if $A$ is a set, then $P(A)$ is the set of boxes that contain elements from $A$.
A: $\{B\}$ is a set with one element; namely, $B$. (The fact that $B$ itself may be a set is irrelevant here). Saying that $\{B\}\in P(A)$ is saying that $\{B\}$ is a subset of $A$ (since $P(A)$ is the set of all subsets of $A$). Now, what does it mean for $\{B\}$ to be a subset of $A$? It means that each element of $\{B\}$ is an element of $A$. But the only element of $\{B\}$ is $B$ itself, which is not an element of $A$ (not in your specific example, nor in most examples that this question contemplates). Rather, the elements of $B$ are elements of $A$.
A: Let's check.  We know $\{B\}=\big\{\{1\}\big\}$ and $P(\{1,2\})=\big\{\emptyset,\{1\},\{2\},\{1,2\}\big\}$.  So:


*

*Does $\big\{\{1\}\big\}=\emptyset$.  No.

*Does $\big\{\{1\}\big\}=\{1\}$.  No.

*Does $\big\{\{1\}\big\}=\{2\}$.  No.

*Does $\big\{\{1\}\big\}=\{1,2\}$.  No.


So $B \not\in P(\{1,2\})$.
Note that $\big\{\{1\}\big\}$ can be informally thought of as "a bag containing a bag containing $1$" whereas $\{1\}$ is "a bag containing $1$".
A: We have $$P(A)=\{C : C \subset A\}.$$
So $$b \in A \Longrightarrow \{b\} \in P(A)$$
 but 
$$\require{cancel}B \subset A \cancel{\Longrightarrow}\{B\} \in P(A)$$ since $\{B\} \notin A,$ however
 $$B \subset A \Longrightarrow \{B\} \subset P(A) \Longrightarrow  \{B\}\in P(P(A)).$$
