I'm currently learning about equivalence relations. I understand that an equivalence relation is a relation that is reflexive, symmetric, and transitive. But I'm having trouble proving the transitive property.
For example:
Suppose we have the relation ~ on the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ by the rule f~g iff there is a $k \in \mathbb{R}$ such that $f(x) = g(x)$ for every $x \geq k$ is an equivalence relation.
$$$$
To prove that it is reflexive I said:
Suppose we have a real number k such that for every $x$, $x \geq k$. Then we can choose any k and f(x) = f(x) for all x. So f~f. Thus the relation is reflexive.
^It seems a little short.. and trivial so I'm wondering if this part is even right.
$$$$
To prove that it is symmetric I said:
Suppose we have two functions such that f(x) = g(x). Then we can similarly say that g(x) = f(x). So it follows that f~g and g~f. Therefore the relation is symmetric
^Again, I'm a little confused about this. It feels too short.
$$$$
Now I'm having trouble proving the transitive property.
Could someone look over my proofs for the reflexive and symmetric properties of the relation and give me a hint on how to approach the transitive property?
Best Answer
In the argument for symmetry, you've shown that if $f = g$ then $f \sim g$ and $g \sim f$. But this does not guarantee symmetry, which requires that if $f \sim g$ then $g \sim f$. To do this, $f \sim g$ shows that there is a constant $k$ such that $f(x) = g(x)$ for all $x \geq k$, and you need to show that there is a constant $l$ such that $g(x) = f(x)$ for all $x \geq l$.
For transitivity, you must show that if $f \sim g$ and $g \sim h$ then $f \sim h$. In your case, this means there are constants $k$ such that $f(x) = g(x)$ for $x \geq k$ and $l$ such that $g(x) = h(x)$ for $x \geq l$, and you need to show that there is a constant $m$ such that $f(x) = h(x)$ for all $x \geq m$. How might you produce such an $m$ in terms of the other data you have available?