Equivalence relation on class of all sets

Question

Let $\mathcal{A}$ be the class of all sets. Prove that "has the same cardinality as" defines an equivalence relation on $\mathcal{A}$.

I know that equivalence relations must be reflexive, symmetric and transitive. So I think the best way to go about this would be to just show that each individual property is satisfied.... but I think I need to do more than just say hey look, this is obvious. And I am not even sure where to start.

Answer 1

Yes, you should just show that each individual property is satisfied

Reflexivity: The identity function works ($1_A: A \to A$).

Symmetry: Given $f: A \to B$ bijective, consider $f^{-1}:B \to A$.

Transitivity: Given $f: A \to B$ and $g:B \to C$, what can you say about the composition $g \circ f: A \to C$?

Answer 2

Often it's difficult to tell exactly when something is supposed to be obvious. The thing to do is almost always to talk to your professor about it.

What she/he might be looking for is something on the level of bijections, which is the way we talk about two sets having the same cardinality.

So, in order to prove that this relation is:

• Reflexive -- construct a bijection from a set $A$ to itself.
• Symmetric -- given a bijection from $B \rightarrow A$, construct a bijection from $A \rightarrow B$.
• Transitive -- given a bijection from $A \rightarrow B$ and one from $B \rightarrow C$, what's the bijection from $A \rightarrow C$?