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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.

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What is $\phi$? Do you mean $\varnothing$ (the empty set)? – Asaf Karagila Feb 27 '14 at 9:48

4 Answers 4

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 $\mathcal P(\{1,2,3\})$?

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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 $B=\mathcal P(A)$.

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I am little confused that what should be $P(\phi)$ . Is it {$\phi$} or {$\phi$ ,{$\phi$}} – jigja Feb 27 '14 at 10:10
Write down the definition of $\mathcal P(\varnothing)$. Then try to see what sets satisfy that definition. – Asaf Karagila Feb 27 '14 at 10:11
still confused... – jigja Feb 27 '14 at 10:30
Write in a comment when a mathematical object is an element of $\mathcal P(\varnothing)$. – Asaf Karagila Feb 27 '14 at 10:35
@jigja - please, remember one of the basic fact about set theory: elements and subsets are different. $\emptyset$ has no elements, but it has subsets, because every set has subsets; at least two : itself and $\emptyset$ (apply the definition of subset to a set $X$ wahtever, to see that $\emptyset$ and $X$ itself are subsets of $X$). And now the final step : apply the previous fact using $\emptyset$ as $X$ ... (You can use the guide of the above comments : if $A$ is finite with $n$ elemnts, the elements of $\mathcal P(A)$ are $2^n$; how many elements has $\emptyset$ ?)... – Mauro ALLEGRANZA Feb 27 '14 at 11:00


If $A$ is finite then $\left|\wp\left(A\right)\right|=2^{\left|A\right|}$. Before using this first ask yourself the question: why?

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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.

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