Relation of equipotence is an equivalence relation?

Question

How do I argue that equipotence relation is an equivalnce relation over power set of $X$?

I have started self-study of real analysis with Royden's book. I don't see what does equipotence relation on a single set X mean? Because equipotence is defined to be the property of two sets A and B, if there is a one-to-one and onto mapping between A and B. Then what does equipotence relation on power set of $X, \mathcal{P}(X)$, would mean? The $\mathcal{P}(X)$ conatains many subsets, does equipotence on $\mathcal{P}(X)$ mean that there is an invertible mapping between any two sets in $\mathcal{P}(X)$, whenever possible?

Answer 1

(Too long for a comment.)

I don't see what does equipotence relation on a single set X mean?

The question is about equipotence being "an equivalence relation over power set of $X$". Nowhere is it mentioned or implied that equipotence would be defined "on a single set X".

Because equipotence is defined to be the property of two sets A and B, if there is a one-to-one and onto mapping between A and B.

In other words, a bijection between $A$ and $B$. That's correct.

Then what does equipotence relation on power set of $X, \mathcal{P}(X)$, would mean?

Precisely what the previous definition says:  for any two sets $\,A, B \in \mathcal{P}(X)\,$ (which is the same as $\,A, B \subseteq X\,$ by the definition of the power set), $A$ and $B$ are said to be equipotent iff there exists a bijection between $A$ and $B$.

The $\mathcal{P}(X)$ conatains many subsets, does equipotence on $\mathcal{P}(X)$ mean that there is an invertible mapping between any two sets in $\mathcal{P}(X)$, whenever possible?

That's worded rather confusingly. Of course not any two subsets of $\,X\,$ are equipotent. But two subsets $A,B$ are indeed equipotent iff there exists a bijection between them.