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 $x^2$.
 A: 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:

*

*Since for every mapping $f$ the equation $f(0)=f(0)$ is trivially true, the relation is reflexive.

*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 $\mathbb{R}$).


*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.

