What is the cardinality of $\mathcal P(\mathcal P(\varnothing))$ where $\mathcal P$ denotes the power set? What is the cardinality of $\mathcal P(\mathcal P(\varnothing))$ where $\mathcal P$ denotes the power set?
I am little bit of confused with it. Help me please.
 A: By definition the power set of any set $X$ is the set of all subsets of $X$, i.e.
$$
\mathcal P(X) = \left\{\, U\subseteq X\,\right\}.
$$
To find $\mathcal P(\varnothing)$, we need to find all the subsets of $\varnothing$. Recall the definition of subsets: A set $A$ is a subset of a set $B$ if and only if every element of $A$ is also an element of $B$, i.e.
$$ A\subseteq B \quad\text{if and only if}\quad \forall x: x \in A \Rightarrow x\in B.$$
To find subsets of $\varnothing$ we have use the definition to obtain
$$ A\subseteq \varnothing \quad\text{if and only if}\quad \forall x: x \in A \Rightarrow x\in \varnothing.$$
Now the definition of $\varnothing$ comes to mind, we know that $x\in \varnothing$ is always false, since $\varnothing$ has no elements. Thus $A$ can't have any elements either, since any element of $A$ had to be an element of $\varnothing$ as well. The only candidate for $A$ is the empty set, $A=\varnothing$. Is it really a subset, i.e. $\varnothing\subseteq\varnothing$? Yes, since "every element of $\varnothing$ is also an element of $\varnothing$" is a tautology since there aren't any elements to begin with. Thus, the only subset of $\varnothing$ is $\varnothing$ itself, so the power set
$$ \mathcal P(\varnothing) = \{\varnothing\}$$
has exactly one element.
In the next step you have to find
$$\mathcal P(\mathcal P(\varnothing)) = \mathcal P(\{\varnothing\}),$$
which consists of all subset of the $1$-element set $\{\varnothing\}$. Again $\varnothing$ is a subset (it always is), but now we have $1$-element in our set, so we can have this element in a subset as well. We conclude that the subsets of the $1$-element set $\{\varnothing\}$ are


*

*the empty set $\varnothing$,

*the $1$-element set $\{\varnothing\}$.


Together we found the power set
$$
\mathcal P(\mathcal P(\varnothing)) = \mathcal P(\{\varnothing\}) = \{ \varnothing, \{\varnothing\}\},
$$
which has $2$ elements, so its cardinality is $2$.
Can you find $\mathcal P(\{1,2,3\})$?
A: HINT: If you are having troubles, work step by step. Calculate what $\mathcal P(\varnothing)$ is, and then calculate $\mathcal{P(P(}\varnothing))$ is. Then count the elements.
If you have theorems connecting the cardinality of $A$ to the cardinality of $\mathcal P(A)$, then apply them. Step by step. $A=\mathcal P(\varnothing)$ and $B=\mathcal P(A)$.
A: Hint:
If $A$ is finite then $\left|\wp\left(A\right)\right|=2^{\left|A\right|}$. Before using this first ask yourself the question: why?
A: A set is called a power set because, if the set is of size n, then any subset either contains an element or doesn't contain it. Thus for any element we have 2 choices, whether to include it or not. That is how we get $2^n$ subsets.
Use this principle.
A: A subset of any set  contains 2^n elements
The power set of an empty set contains 2^0 elements=since an emty set has no element.
|P(∅)|=2^0=1
P(∅)={∅} since the empty set is a subset of any set
