If $A^X$ is equipotent to $B^X$ does it imply that A is equipotent to B?

Question

I was able to prove the converse of this statement to be true.However, I have some doubt whether the statement above is true. If true please provide a rigorous proof and if false please provide a counter-example or atleast a reason for the falsity of the statement.

Answer 1

Suppose that $X$ is a nonempty finite set. Let $A$ and $B$ be sets such that $\left|A^X\right|=\left|B^X\right|$. We consider two cases. First, suppose that $A$ is finite. Then, $$|B|\leq |B|^{|X|}=\left|B^X\right|=\left|A^X\right|=|A|^{|X|}<\infty\,,$$ whence $B$ is also finite. Thus, $$|B|^{|X|}=|A|^{|X|}\text{ implies }|B|=|A|\,.$$ Now, we assume that $A$ is infinite. Then, it is well known (easy to prove) that $\alpha^n=\alpha$ for all infinite cardinal numbers $\alpha$ and for any positive integer $n$. Thus, $$|A|=|A|^{|X|}=\left|A^X\right|=\left|B^X\right|=|B|^{|X|}=|B|\,.$$

Finally, we drop the condition that $X$ is a nonempty finite set. If $X=\emptyset$, then $$A^X=\{\emptyset\}=B^X$$ for any sets $A$ and $B$. If $X$ is an infinite set, then $$\left|A^X\right|=2^{|X|}=\left|B^X\right|$$ for all countable sets $A$ and $B$ with at least two elements.

Answer 2

No, $\{0,1\}^\mathbb{N}$ has the same number of points as $\{0,1,2\}^\mathbb{N}$ but $\{0,1\}$ and $\{0,1,2\}$ do not.