1. ## function and relation

The examples of equivalence classes in my book are mostly related to integers. I would like to know this:

Can a continuous function have equivalence classes?

2. What conditions need to be satisfied for a continuous function to be:

1) A well-defined function
2) An equivalence relation?

Answer those and you will get your result; note that you're not even considering the continuity.

3. Do you mind showing a simple example for someone like me?

4. Sure:

Any function is a relation. Specifically, a function $f:A \to B$ is a relation $f \subseteq A \times B$, such that for any $a \in A$, there is exactly one $b \in B$ such that $(a,b) \in f$.

Now, for f to be an equivalence relation, its domain has to be its range - $f: A \to A$.

Specifically, it also has to be reflexive. That is, for each $a \in A, \ (a,a) \in f$.

What does that tell you about f?

5. Very interesting. It looks like an identity function, but what does the equivalence classes look like since the points are so dense?

6. Oh, yeah, you already said about the continuity being irrelevant. Thank you for your valuable time.

7. Well, what I meant was that you can get the result without assuming continuity, but the function you get (identity function) is always continuous. The equivalence class of $x \in \mathbb{R}$ is $[x] = \{ y \in \mathbb{R} : (x,y) \in f \}$ $= \{ y \in \mathbb{R} : f(x)=y \} = \{ y \in \mathbb{R} : x = y \} = \{ x \}$

So each point's equivalence class is itself.

8. Originally Posted by Defunkt
The equivalence class of $x \in \mathbb{R}$ is $[x] = \{ y \in \mathbb{R} : (x,y) \in f \}$ $= \{ y \in \mathbb{R} : f(x)=y \} = \{ y \in \mathbb{R} : x = y \} = \{ x \}$

So each point's equivalence class is itself.
Aha, that's what I was hoping to see.

Beautify! Thanks again.