I am currently stumped with the following question regarding power sets. I don't know how to approach it.
Find the cardinality of |$\mathcal{P}(\mathcal{P}(\mathcal{P}(A)))$|
I am aware that the subsets of a power set will take the form of $2^n$. I however, don't know how to compute/ approach this problem. If I attempt to tackle it from the inside or the outside I still fail to get to correct answer which is ... $2^{(2^{2^{m}})}$
Edit: Thank You guys for all the answers I got the idea.
$\endgroup$ 23 Answers
$\begingroup$We have:
$$ |\mathcal{P}(\mathcal{P}(\mathcal{P}(A)))| = 2^{|\mathcal{P}(\mathcal{P}(A))|} = 2^{2^{|\mathcal{P}(A)|}} = 2^{2^{2^{|A|}}} = 2^{2^{2^{m}}} $$
$\endgroup$ $\begingroup$If $A$ has $n$ elements, then the power set of $A$ has $2^n$ elements as you said. But that means that the power set of the power set of $A$ has $2^{2^n}$ elements by the same argument. Repeating that once more, we get $| \mathcal{P}(\mathcal{P}(\mathcal{P}(A))) | = 2^{2^{2^n}}$.
$\endgroup$ $\begingroup$Recall that the powerset of a set with $n$ elements is of size $2^n$. That is to say:
$$|\mathcal{P}(X)|=2^{|X|}$$
or, using more suggestive notation, letting $2^X$ notate the powerset of $X$ we have
$$|2^X|=2^{|X|}$$
Now, the nested powerset, we have:
$$|\mathcal{P}(\color{red}{\mathcal{P}(\mathcal{P}(A))})|=2^{|\color{red}{\mathcal{P}(\mathcal{P}(A))}|}$$
Note that the red $\color{red}{\mathcal{P}(\mathcal{P}(A))}$ is itself a set, albeit a slightly complicated one compared to what you might be used to using. Its properties are otherwise the same and it can appear in identities just like any other. We arrived at the above by using it in place of $X$ that we originally wrote the identity with. Next, we can continue to expand what is in the exponent. In doing so we eventually arrive at a final answer of
$$2^{(2^{(2^n)})}$$
$\endgroup$
