Prove a set is a Partition The question: Let $r \in \mathbb{R}$. Define the set $A_{r} = \{ (x,y) \in \mathbb{R} \times \mathbb{R} \mid x^{2} + y^{2} = r^2 \} $. Prove $\{ A_{r} \mid r \in \mathbb{R} \}$ is a partition of $\mathbb{R} \times \mathbb{R} $. 
The proof: 
Let $(x,y) = (r,0) $. Then, $x^2 + y^2 = r^2 + 0^2 = r^2$. So, $(r, 0) \in A_{r}$ and the set is non-empty.
Let $r,s \in \mathbb{R}$ such that $r \neq s$. Suppose by contradiction, that $A_{r} \cap A_{s} \neq \emptyset$. This implies that  $(x,y) \in A_{r} \cap A_{s}$. So, $x^2 + y^2 = r^2$ and $x^2 + y^2 = s^2$. But, $r^2 = s^2$. This is where I am stuck in this proof. I said that a circle with center at the origin,  cannot have two radii. However I think that this is not way to end this step in the proof. I also said that $ r^2 - s^2 = 0$, which by factoring means $(r-s)(r+s) = 0$, which means that $r = -s$. Am I on the right track? 
Thank you very much! 
 A: Note that $A_{r} = \{ (x,y) \in \mathbb{R} \times \mathbb{R} |x^{2} + y^{2} = r^2 \} =\{ (x,y) \in \mathbb{R} \times \mathbb{R} |x^{2} + y^{2} = (-r)^2 \}=A_{-r}$.
So it suffices to prove that $\mathbb{R} \times \mathbb{R}=\{A_r \;|\; r\geq0,r\in \mathbb{R} \}$.
You can define an relation by 
$$(x_1,y_1)\sim (x_2,y_2)\iff x_1^2+y_1^2=x_2^2+y_2^2$$
You can verify that this is an equivalence relation.
So it partition $\mathbb{R} \times \mathbb{R}$ into the equivalence classes that are $A_r$ where $r\geq 0$
A: To prove that a set $\mathcal{P}$ is a partition, you need to prove (among other things) that if $A,B\in\mathcal{P}$ and $A\neq B$, then $A\cap B=\emptyset$.  Notice that this is different from what you're trying to prove: you're assuming that $A_r,A_s\in \{ A_{r} | r \in \mathbb{R} \}$ and $r\neq s$.  What you should be assuming instead is that $A_r\neq A_s$, since it may be possible that $A_r=A_s$ even if $r\neq s$.  So in your proof by contradiction, your goal is not to find that $r=s$ (which as you observed need not be true since you could have $r=-s$), but instead to find that $A_r=A_s$.
