Question

f(×) ~ g(×) if f'(×)=g'(×) show that ~ is an equivalence relations
​

Answer 1

Step-by-step explanation:

Questions Tags Users Badges Ask

Up vote

0

Down vote

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

equivalence-relations

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.

share improve this question follow

asked

Mar 23 at 14:52

idkkkkkkk

3●33 bronze badges

2

Welcome to MSE. What did you try? Where are you stuck? – José Carlos Santos Mar 23 at 14:59

Here's a MathJax tutorial :) – Shaun Mar 23 at 15:12

add a comment

2 Answers

order by

Up vote

0

Down vote

Reflexivity: for some function f we have f(x)=f(x) for every f∈R. Hence f(x)≠f(x) for 0 points i.e. for finitely many. Hence f∼f.

Symmetry: for two functions f,g, for which f(x)=g(x) for all but finitely many points (i.e. f∼g), we also have g(x)=f(x) exactly those points, i.e. also for finitely many points (so g∼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∈R|f(x)≠g(x)}∪{x∈R|g(x)≠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∼h.