# Is this relation an Equivalence relation?

{(A,B) : A, B ⊆ X, there is a bijective f : A → B}, X is limited.

I have to show if this is (for proving that's an equivalence relation):
$R ⊆ X \times X$
I) reflexive (if $∀x ∈ X : (x, x) ∈ R$)
II) symmetrical (if $∀x,x' ∈ X : (x, x') ∈ R ⇒ (x,x') ∈ R$)
III) transitive (if $∀x, x', x'' ∈ R : (x, x') ∈ R (x',x'') ∈ R ⇒ (x, x'') ∈ R$)

Alright, I understand what and how I have to do it, but somehow not with this task.

Maybe someone could show me how to transform the very first line into something I can work on. I'm also a little confused about that $R⊆X \times X$ thing. It's A, B ⊆ X and I'm afraid that my 3 points I have to show can't be applied that easily on my task.

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The relation is on the set of subsets of $X$, in fact it is a subset of $\mathcal{P}(X)\times\mathcal{P}(X)$. –  egreg Apr 26 '13 at 13:05
@egreg So would it be R ⊆ A x B in this case? –  nullmoon Apr 26 '13 at 13:07
No, $R\subseteq\mathcal P(X)\times\mathcal P(X)$, and if $A,B\subseteq X$ (that is, $A,B\in\mathcal P(X)$), then $(A,B)$ might be an element of $R$. –  Berci Apr 26 '13 at 13:08

For this particular relation $R$, we have two base sets, one is $X$, its subsets are the 'elements' among which $R$ is interpreted, so the base set of $R$ now is the power set $\mathcal P(X)$, that is, $R\subseteq \mathcal P(X)\times\mathcal P(X)$.
You need to show that, for all $A,B,C\in\mathcal P(X)$:
1. $(A,A)\in R$, i.e. there is a bijection $A\to A$.
2. If there is a bijection $A\to B$, then there is one $B\to A$.
3. If there are bijections $A\to B$ and $B\to C$, then there is one $A\to C$.
Sets as $A=\{1,2\}$ to show 'something like this'.. What do you mean? –  Berci Apr 26 '13 at 14:08