How do you solve for the cardinality of a power set of some complex set? (i.e. $|\mathcal P(A^n)|$ , $|\mathcal P(A\cup B)|$ )

Question

Suppose $A$ is some set such that $A = \{a_1,a_2,\dotsb,a_n\}$.

We know that $|A|=n$.

We know that $\mathcal P(A)= 2^n$.

Now let $A^n$ denote the cartesian product of a set A with itself n times. $$\text{(i.e.) }\left[A^n=\underbrace{A\times A\times\dotsb\times A}_\text{$n$}\right]$$

We know the cartesian product of a sets $A\times B=|A|*|B|$

Thus $|A^n| = \left[\underbrace{|A|* |A|*\dotsb*|A|}_\text{$n$}\right] = |A|^n$

??? $|\mathcal P(A^n)|=\dots$

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Question:

• What is a general way to solve for the cardinality of a power set of some complex set equation?
• If there are general steps for finding the cardinality of a power set of some complex set equation, what are they?
• Is $|\mathcal P(A^n)| = 2^{n^n} =2$^n^n?

Answer 1

You already know that $|\mathcal P(X)|=2^{|X|}$.

If $X=n^n$ then $|\mathcal P(X)|=2^{n^n}$. And indeed if $|A|=n$ and $X=A^n$, then $|X|=n^n$.

Generally speaking, finding the cardinality of the power set requires you to find the cardinality of $X$, if $X$ is a union, or product or whatever, then you need to calculate the cardinality of this $X$, and take $2^{|X|}$.

Answer 2

If you think of any subset of a set $Y \subseteq X$ as a binary string, where each character represents an element of $X$ and is $1$ or $0$ if the element is in $Y$ or is not in $Y$ respectively, you find out that all the distinct binary strings of "length" $|X|$ are in 1-1 correspondence with all the distinct subsets of $X$. We know the elements of $ \mathcal P \left({X}\right) $ are all the subsets of $X$, and we also know the number of binary strings of "length" $|X|$ is $2^{|X|}$, so: $$ |\mathcal P \left({X}\right)| = 2^{|X|} $$ For any set $X$ (finite or not).