Question

f ~ g <-> {x in N | f(x) not equal g(x)} is finite.
show that this is an equivalence relation.
HELP!!! especially for the reflexive/transitive part

Answer 1

f ~ f : {x in N | f(x) not equal f(x)} = {} and an empty set is finite. therefor f ~ f

Answer 2

can you help with symmetry and transitivity too?? thank you SO much!

Answer 3

symmetry: assume f ~ g
{x in N | f(x) not equal g(x)} is finite

which we can also write {x in N | g(x) not equal f(x)} is finite
or g ~ f

Answer 4

transitivity: ASSUME
f ~g OR {f(x) not equal g(x)} is finite
g~h OR {g(x) not equal h(x)} is finite

PROVE:
f ~ h OR {f(x) not equal h(x)} is finite
hmm... kinda hard to figure

Answer 5

that's where i got stuck too, since it is not evident f(x) and h(x) are not equal...

Answer 6

but thank you for your input, i will look into it some more.

Answer 7

we have to assume that f(x) is not equal to h(x) and prove that the set is finite.