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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.

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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. $f \sim h$.

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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.

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