Equivalence and partition question ,need help with reflexive property

Question

Let R be the set of the real numbers. Prove $\{G_r\}_{r\in\mathbb{R}}$ is a partition of $\mathbb{R}^{2}$

Let $G_r= \{(x,y) :x^2+y^2=r\}$. For each $r∈R$

I have to do 3 things:

1. Prove it is a partition.

2. Find it’s equivalence class.

   3.describe geometrically.

I done 3.

2. Done by Mr Diaz To prove 1, I have to prove that it’s subsets are disjoint and form G, correct ?

Given G_r Is it enough to do this:

Prove that given a partition $\mathcal{P}$ of a set $A$ nonempty, there exists a unique equivalence relation on $A$ from which it is derived

My dilemma the problem is set up in order to prove P without referring to 2?

So, if it is reflexive, then $x \sim x$ implies that $x-x=0$ so that $x^2 + y^2=x^2+y^2$ implies that $x^2-x^2=y^2-y^2=r-r=0

I have no problem with the other properties.

“~” equivalent to

I have another attempt which makes no sense, so I won’t write. This is my best attempt...

When Pinter stated for each $r∈R$ I thought I could assume it could mean for all of them

Answer 1

I've seen the exercise of the book, you need to find the equivalence relation that corresponds to the partition.

Define this relation: Let $(x_{1}, y_{1}), (x_{2}, y_{2})\in \mathbb{R}^{2}$, then $(x_{1}, y_{1})\sim(x_{2}, y_{2})$ iff $x_{1}^{2} + y_{1}^{2} = x_{2}^{2}+ y_{2}^{2}$

It is reflexive on $\mathbb{R}^{2}$, since $x_{1}^{2} + y_{1}^{2} = x_{1}^{2}+ y_{1}^{2} \leftrightarrow (x_{1}, y_{1})\sim(x_{1}, y_{1}), \forall (x_{1}, y_{1})\in \mathbb{R}^{2}$