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**1. Homework Statement**

**2. Homework Equations**

I don't think there are any in this case

**3. The Attempt at a Solution**

I know that in order to prove R is an equivalence relation, I'd have to show that it is Reflexive, Symmetric, and Transitive. I'm not sure why, but I'm finding this a bit difficult in this case, possibly because the "elements" in question are now sets instead of just regular elements.

Here's what I've thought up so far:

For Reflexivity, I have to demonstrate that ##\forall S\in A((S,S) \in R)##. Since ##A## is the class of all possible sets, of course ##S \in A##, which gets the first requirement out of the way.

As for ##[\exists f : S \rightarrow S \space\ bijective function]##... I can think of one possible example for such a function, namely the identity function ##f(x) = x##; since N maps into itself, the identity function should be both injective (each element in the domain outputs itself in the codomain) and surjective (each element in the codomain is output by itself).

I'm not sure if this is right, though.

I'm not sure how to demonstrate symmetry here. By definition I'd have to show ##\forall S \forall Q ((S,Q) \in R \rightarrow (Q,S) \in R)##, which really just boils down to showing that ##[\exists f : (S \rightarrow Q \space\ bijective function) \rightarrow (Q \rightarrow S \space\ bijective function) ]##, if I'm not mistaken?

However, in this case, S and Q are two different functions, so I can't just use the identity function like with reflexivity... and that's where I'm stuck. I'm not sure how to show symmetry or transitivity here.