Not a precise question on equivalence class.

Question

Consider $f:X\longrightarrow Y$. Define a relation $\sim$ on $X$ by $a\sim b$ iff $f(a)=f(b)$. I proved that $\sim$ is an equivalence relation and that if $f$ is onto and $X/\sim$ is the set of equivalence classes, then there is a bijective function between $X/\sim$ and $Y$. Now comes the odd question, which I don't know if it is a "famous" question or exercise:

If $X=Y=\mathbb R$, describe a geometric method that allows you to visualize the equivalence class of a real number.

I don't know what this means. Is there any interpretation you could give or an example of a "geometric method" that could be used for such means?

I'm sorry if the question on the book is not very specific, I just want to know if any of you could give some sense to that question. Thanks.

Answer 1

In this case, the equivalence class of $x$ is the set $\{z \in \mathbb R \mid f(z) = f(x) \}$ for your function $f$. This can be found by drawing the graph of $f$ and checking for which values of $z$ the graphs $f$ and $y = f(x)$ intersect (for the given $x$).

Answer 2

Draw the graph of $f$. Then $a \sim b$ if and only if $f(a)$ and $f(b)$ are at the same height.

Therefore, if you draw a horizontal line at the point $(a,f(a))$ the class of $a$ is "exactly" the intersection between this line and the graph.

"exactly" means that the class consists of all the $x$ -coordinates of the intersection points, or equivalently of the the projection of the intersection onto the $x$ axis.

Answer 3

$$x_1 \sim x_2 \iff f(x_1) = f(x_2) = k \iff x_1,x_2 \in f^{-1}(\{k\})$$

Two numbers are in the same class if they are in the pre-image of the same number. Consider the graph of the function. Take a horizontal line $y = k$. It will cut the graph in some points, project them in the $x-$axis. These projections are in the same class.