An equivalence relation on $\mathbb{R}^2\setminus \{(0,0)\}$ (Resolved)

Question

So this is from one of my homework problems in which we consider the map $f:\mathbb{R}^2\setminus \{(0,0)\} \to \mathbb{R}^2\setminus \{(0,0)\}$ where $f(x,y)=(2x,\frac{1}{2}y)$. Then it goes like "let $\sim$ be the equivalence relation generated by $(x,y)\sim f(x,y)$". However, I don't see why $\sim$ is an equivalence relation since we don't have any of reflexivity, symmetry or transitivity here.

More precisely, reflexivity fails since $(x,y)\neq f(x,y)$ whenever $(x,y)\in \mathbb{R}^2\setminus \{(0,0)\}$.

To check whether we have symmetry for $\sim$, notice that $$f(x,y)\sim (x,y)\iff (x,y)=f^2 (x,y)\iff f \text{ is an involution}$$ the last part of which does not hold.

For transitivity we can only deduce that $f^k (x,y)\sim f^{k+1} (x,y)$ for all integers $k$. But it is not the case that e.g. $f(x,y)\sim f^3 (x,y)$.

Can someone explain what went wrong? Thanks!

Answer 1

Let $\sim$ be the equivalence relation generated by $(x,y)\sim f(x,y)$.

It means that $\sim$ is the smallest equivalent relation that contains the given relation.

And yes, in effect, it means that we have to add pairs to our relation in order to get it reflexive, symmetric and transitive.

So, for a specific example, we will have e.g. $ (12,5)\sim (3,20)$.

Answer 2

For every set of equivalence relations, their intersection (i.e. the relation that is valid when all those relations are valid) is again an equivalence relation. It follows that, taken an arbitrary relation $R$ on a set $X$, you can take the intersection of all equivalence relations on $X$ containing $R$. This intersection will still be an equivalence relation, and it will still contain $R$, and so it will be the smallest equivalence relation on $X$ containing $R$. We say that this is an equivalence relation generated by $R$.

Rewinding back to your problem: you are not given an equivalence relation, you are expected to find and work with the equivalence relation generated by your relation. I suspect that this equivalence relation is given by:

$$(x,y)\sim(z,t)\Longleftrightarrow (\exists i\in\mathbb Z)(z=2^ix \land t=2^{-i}y)$$