# Determining if this set is an equivalence relation verification

I am asked to show the follow set if an equivalence relation on a given set.

So I have:

Set: A=Map$$(\mathbb{R},\mathbb{R})$$
Relation: $$f \sim g$$ if $$f(0)=g(0)$$ or $$f(1)=g(1)$$.

I know that this is not an equivalence relation but I was wondering if someone could check my reasoning.

Reflexivity: $$f \sim f$$ implies $$f(0)=f(0)$$ or $$f(1)=f(1)$$

This will always be true for any function in the real numbers so the reflexivity property is always satisfied.

Symmetry: $$f \sim g$$ and $$g \sim f$$ implies $$g(0)=f(0)$$ or $$g(1)=f(1)$$

The equality still preserves the original definition so this is essentially the same as the original statement so the symmetry property is still satisfied.

Transitivity: $$f \sim g$$, $$g \sim h$$ implies $$f \sim h$$ implies that we should have:

$$f \sim g:$$ $$f(0)=g(0)$$ or $$f(1)=g(1)$$

$$g \sim h:$$ $$g(0)=h(0)$$ or $$g(1)=h(1)$$

implies we should have:

$$f \sim h:$$ $$f(0)=h(0)$$ or $$f(1)=h(1)$$.

Let $$f(x)=1$$ and $$g(x)=x$$ and $$h(x)=2$$

The $$f \sim h:$$ $$f(0)=h(0)$$ or $$f(1)=h(1)$$ fails so this is not an equivalence relation. Is that the right idea?

EDIT: I changed $$h(x)=2$$ instead of $$x^2$$.

• One comment: I don't like the way you stated the reflexivity. IMO a correct way of writing it would be " $\forall f: f$ ~ $f$ holds because $f(0)=f(0)$ (and $f(1)=f(1)$) trivially" Commented Jun 3, 2021 at 23:44
• Fair enough! Thanks a lot. Commented Jun 4, 2021 at 5:36
• I'm amazed because nobody has helped you yet ! You have the right ideas (though your counterexample is not completely right), if you want I can share with you the way I would show it's not an equivalence relation in an answer below. Commented Jun 4, 2021 at 14:39

You are right, the relation on your exercise is not an equivalence relation because it's not transitive. You have the right ideas, but I see that you are having troubles to explain them. The implication

$$f \sim f \Rightarrow f(0)=f(0) \vee f(1)=f(1)$$

is not what you need to prove to state that the relation is reflexive (we already know that if $$f$$ happens to be related with itself, it's because $$f(0)=f(0) \vee f(1)=f(1)$$ because of the definition of being related). Instead, what you are interested in proving is that the statement

$$\forall f: f \sim f$$ is true.

Below I show you how I would solve your exercise:

1. Since for every mapping $$f$$ the equation $$f(0)=f(0)$$ is trivially true, the relation is reflexive.
2. Let $$f$$ and $$g$$ be two maps for which the statement $$f$$ ~ $$g$$ is true. WLOG let's suppose that $$f(0)=g(0)$$. Then we have $$g(0)=f(0)$$ too, and $$g$$ ~ $$f$$ holds.

(note that points 1 and 2 are true because equality is an equivalence relation in $$\mathbb{R}$$).

1. Let's show that the relation is not transitive. Consider the mappings $$f(x)=1$$, $$g(x)=x$$, $$h(x)=0$$. Since $$f(1)=g(1)$$ and $$g(0)=h(0)$$, we have that $$f$$ ~ $$g$$ and $$g$$ ~ $$h$$. However, $$f$$ is not related with $$h$$, since $$f(0) \neq h(0)$$ nor $$f(1) \neq h(1)$$. We have found mappings $$f$$, $$g$$ and $$h$$ for which $$f \sim g$$ and $$g \sim h$$, but $$f \nsim h$$, so the statement $$f \sim g \wedge g \sim h \Rightarrow f \sim h$$ is false, and the relation is not transitive, thus it's not an equivalence relation.