# Show that ∼ is an equivalence relation for f ∼ g if f(x) = g(x) except at possibly finitely many points

Let S denote the set of functions f : R → R and define a relation ∼ on S by declaring f ∼ g if f(x) = g(x) except at possibly finitely many points (that is, the set of x where f(x) = g(x) is either empty or finite). Show that ∼ is an equivalence relation.

I know that to show an equivalence relation you must show that it is reflexive, symmetric and transitive but I am confused as to how to do it for this problem. Any help would be appreciated.

• Welcome to MSE. What did you try? Where are you stuck? Mar 23, 2020 at 14:59
• Mar 23, 2020 at 15:12

## 2 Answers

Reflexivity: for some function $$f$$ we have $$f(x) = f(x)$$ for every $$f \in \mathbb{R}$$. Hence $$f(x) \neq f(x)$$ for $$0$$ points i.e. for finitely many. Hence $$f \sim f$$.

Symmetry: for two functions $$f,g$$, for which $$f(x) = g(x)$$ for all but finitely many points (i.e. $$f \sim g$$), we also have $$g(x) = f(x)$$ exactly those points, i.e. also for finitely many points (so $$g \sim f$$)

Transitivity: For three functions $$f,g,h$$ and $$f(x) = g(x)$$ for all but finitely many points and $$g(x) = h(x)$$ for all but finitely many points, then we have $$f(x) = g(x) = h(x)$$ for all those points, for which the above two equalities hold. That is, they hold for points which are not in $$\{x \in \mathbb{R} \vert f(x) \neq g(x)\} \cup \{x \in \mathbb{R} \vert g(x) \neq h(x)\}$$. By assumption, both of those sets are finite and thus their union is also finite. Hence we have $$f(x) = h(x)$$ for all but finitely many points i.e. $$f \sim h$$.

The two first points are quite tautological once rephrased:

Reflexivity: take a function $$f:\mathbb{R}\to\mathbb{R}$$. Do you have $$f(x)=f(x)$$ except at possibly finitely many points (eapfmp)?

Symmetry: take two functions $$f:\mathbb{R}\to\mathbb{R}$$ and $$g:\mathbb{R}\to\mathbb{R}$$. If you have $$f(x)=g(x)$$ eapfmp, do you have $$g(x)=f(x)$$ eapfmp?

You will have to work in order to get the third point:

Transitivity: Take three functions $$f,g,h$$. Suppose that $$f(x)=g(x)$$ eapfmp, and suppose that $$g(x)=h(x)$$ eapfmp (which are not related to the first ones). You have to show that $$f(x)=h(x)$$ eapfmp. Start by naming those two sets of points, and try to show the statement.