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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.

2 Answers 2

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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.

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    Thank you for answering! Ok so, I would like to ask the following, why should I take into account the preimage of the function? I mean, if I would have to explain it to someone how would I say it? I like to think it in that way since it helps me a lot to understand concepts like these. – 2017-01-16
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    If I were to describe it to a non-technical person, I'd probably say the following, though not in exactly these words: a function can be viewed as a labeling of the elements in its domain. We can then decide that two elements are equivalent if they have the same label. If you then consider the set of all elements that have a given label which will clearly be an equivalence class, then that turns out to be exactly the preimage of the labeling. Alternatively, you can think of a function as picking out an attribute, e.g. hair color, that you then group by the groups being equivalence classes. – 2017-01-16
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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.

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    Thank for answering! Well, for the first question, I understand it as implementing constraints such that it covers the whole circle. Perhaps a function with two cases? For y>0 and y<0 or smthng like that? For the second one, could you please guide me through it? I have not done played with equivalence relations before... :/ – 2017-01-15
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    @Mitscaype It's just a plain fact that some relations are not functions. In fact a function is a special type of relation. The circle fails the vertical line test; end of story. In Calc 3, when we make graphs of two-variable functions we often start with level curves, which is really a contour map of the mountain range specified by the function. Google Image "level curves" for a number of pictures. – 2017-01-16
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    Ok, I understand it now. I googled it a little bit. But still, how could I at least start with the second question? Where should I look? – 2017-01-16
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    @Mitscaype Any calculus text would have explanation and nice pictures. Paul's online notes has a page: http://tutorial.math.lamar.edu/Classes/CalcIII/MultiVrbleFcns.aspx – 2017-01-16