Relations Problem 4

Reminder of how we defined the relation:

$$f \thicksim g$$ if and only if there is a $$k \in \mathbb{R}$$ such that f(x) = g(x) for every $$x \geq k$$.

Solution

Reflexive

Let f be a function from the reals to the reals. Pick k=0. (Any value will work.) Then f(x) = f(x) for every $$x \geq k$$. So $$f \thicksim f$$.

Symmetric

Let f and g be functions from the reals to the reals. Suppose that $$f \thicksim g$$. Then, by the definition of $$\thicksim$$, there is a real number k such that f(x) = g(x) for every $$x \geq k$$. But then g(x) = f(x) for every $$x \geq k$$. So we must have $$g \thicksim f$$.

Transitive

Let f, g, and h be functions from the reals to the reals. Suppose that $$f \thicksim g$$ and $$g \thicksim h$$.

By the definition of $$\thicksim$$, this means that there are real numbers k and m such that such that f(x) = g(x) for every $$x \geq k$$, and g(x) = h(x) for every $$x \geq m$$.

Let n = max(k,m). Then f(x) = g(x) for every $$x \geq n$$, and also g(x) = h(x) for every $$x \geq n$$. So f(x) = h(x) for every $$x \geq n$$. So we must have $$f \thicksim h$$.

Since our relation is reflexive, symmetric, and transitive, it is an equivalence relation.

Variant solutions

There's not much scope for variation in this proof. Equivalence relation proofs tend to be largely mechanical application of standard techniques.