What is the cardinality of the power set $P(A \cup B)$

Question

Let $A = \{1, 3, 5\}$ and $B = \{3, 4, 5\}$ be sets.

What is the cardinality of the power set $P(A \cup B)$?

If i'm not mistaking isn't it all the possible combination of these two: $\{\}, \{1,3\}, ...$.

Answer 1

$A \cup B = \{1,3,4,5\}$. The cardinality of the power set of a set of order $n$ is $2^n$. You should be able to calculate the cardinality then.

Answer 2

$A\cup B=\{1,3,4,5\}$, so $$P(A\cup B)=\{\varnothing, \{1\},\{3\},\{4\},\{5\},\{1,3\},\{1,4\},\{1,5\},\{3,4\},\{3,5\},\{4,5\},$$

$$\{1,3,4\},\{1,3,5\},\{1,4,5\},\{3,4,5\},\{1,3,4,5\}\}$$

I provided the explicit power set to let you know what it is. Now you can count how many elements it has.

But you don't have to write out the power set to know its cardinality. $$|P(X)|=2^{|X|}$$

To prove it, consider strings of bits of length $|X|$: $$\underbrace{00\cdots 00}_{|X|},\, \underbrace{00\cdots 01}_{|X|},\ldots,\, \underbrace{11\cdots 11}_{|X|}$$

There are $2^{|X|}$ such unique strings. We can imagine they correspond to all the subsets of $X$ as follows:

Each bit corresponds to the respective element in $X$ in a way that if the bit is $0$, the element is not in the subset, and if the bit is $1$, the element is in the subset.