Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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

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

share|cite|improve this answer

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 $B=\mathcal P(A)$.

share|cite|improve this answer
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?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.