Equivalence Relations of the Circle $x^2+y^2=r^2$

Question

I have been studying Set Theory and this question came up:

Let us assume the set $C_r$, consisting of the points of the circumference of the circle with a center at $(0,0)$ and a radius $r>0$, as a collection of ordered pairs of elements characterized by a binary relation at $\mathbb{R}$ defined as: \begin{equation} (x,y) \in C_r, \quad \text{if and only if} \quad x^2+y^2=r^2 \end{equation}

• How is it possible to define the domain or the image of this relation in order for it to be a function of one variable, that being $x$ or $y$?
• Considering now that $r$ is a variable, define an equivalence relation in $\mathbb{R}^2$ such that the sets $C_r$ and the unit set $\{ (0,0) \}$ to be equivalence classes of the equivalence relation.Could this procedure be generalized for any $\phi: \mathbb{R}^2 \to \mathbb{R}$ and not just for $\phi(x,y)=x^2+y^2$? If yes, what would the correspoding equivalence classes be?

For the first one I think that solving for one of $x$ or $y$ would do the trick, e.g: $x=\pm \sqrt{r^2-y^2}$, but then, this is not a function. I would have to constraint the domain to the semicircle but that will not cover the whole circle. For the second question I am not really sure how to proceed.

Thank you.

Answer 1

You are correct for the first question and B. Goddard elaborates on that.

The second question is "yes". Note that $C_r = \phi^{-1}(r^2)$, the inverse image of $r^2$ with respect to $\phi$. This is what B. Goddard is calling "level curves", but that's an overly geometrical picture. It's simply the set $\{(x,y)\in\mathbb{R}^2\mid\ \phi(x,y)=r^2\}$. This is an equivalence class of the equivalence relation $$(x_1,y_1)\sim(x_2,y_2) \iff \phi(x_1,y_1)=\phi(x_2,y_2)$$ which you can easily prove is an equivalence relation, i.e. that it is reflexive, symmetric, and transitive. Every function will give rise to such an equivalence relation on its domain.

Answer 2

The answer to the first question is "it's not possible." For every $y$ there are two $x$'s that satisfy the equation. A function allows only one occurrence of each value of $x$. (And mutatis mutandis.)

I think the answer to the second question is "yes" (if I understand it.) You're really taking the level curves of the function $\phi$ as equivalence classes, and that should work just fine.