Power set, belongs or inclusion? My doubt is relative to the power set and the notation.
If we have a set $A = \{a,b,c\}$, its power set is $$\mathcal{P}(A) = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},A\}$$
So we get rightly these notations
$a \in A$, $\{a\} \subset \mathcal{P}(A)$, $\{a\} \in \mathcal{P}(A)$?
Thanks in advance!
 A: You already wrote down correctly what $\mathcal{P}(A)$ is. So you can just check manually.


*

*Definitely $a \in A$, because that is how you defined $A$.

*For $\{a\} \subset \mathcal{P}(A)$ to hold, we must have that every element of $\{a\}$ is an element of $\mathcal{P}(A)$. So that would mean that $a \in \mathcal{P}(A)$, but as you can see $a$ never appears as an element of $\mathcal{P}(A)$. So $\{a\} \not \subset \mathcal{P}(A)$.

*Finally, $\{a\}$ does appear as an element in $\mathcal{P}(A)$, so indeed $\{a\} \in \mathcal{P}(A)$.


When getting confused about notation, it is often best to just write out exactly what the notation means according to its definition.
A: We have $a\in A$ and also $\{a\}\in\mathcal P(A)$.
Further we will have $\{a\}\subseteq\mathcal P(A)$ if and only if $a\in\mathcal P(A)$.
Looking at $\mathcal P(A)$ we conclude that this occurs if and only if at least one of the following conditions is satisfied:


*

*$a=\varnothing$

*$a=\{b\}$

*$a=\{c\}$

*$a=\{b,c\}$
Accepting the axiom of regularity we excluded on forehand that the conditions $a=\{a\},a=\{a,b\},a=\{a,c\}$ and $a=A=\{a,b,c\}$ are satisfied. 
This because this axiom "forbids" that $a\in a$.
Note that this concerns special cases. In general if $A=\{a,b,c\}$ then it is not true that $\{a\}\subseteq\mathcal P(A)$. The answers of José and Mark are in that context.
A: Yes, $a\in A$, and, yes, $\{a\}\in\mathcal P(A)$. But it is not true that $\{a\}\subset\mathcal P(A)$.
