# Proof of f(x) = g(x) for all x in R

• C.E

#### C.E

1. Suppose that f and g are continuous functions defined on R and every interval
(a, b) contains some point y with f(y) = g(y). Show that f(x) = g(x) for every x in R.

3. I can show that between any two points in are there is some x such that f(x)=g(x). Is that enough? I don't think it is but have no idea what else I can do. Any ideas?

Let x be any point in R and consider the sequence of nested intervals I_n:= ]x-1/n,x+1/n[ "converging" to {x}. Try to use the hypothesis that every interval (a, b) contains some point y with f(y) = g(y) to find a sequence {x_n} with x_n in I_n such that f(x_n)=g(x_n) and x_n-->x. How does this imply the desired conclusion?

The thing you show is not enough - it is just a restatement of the hypothesis. You've not used the fact that f and g are continuous - that should concern you.

You want to show that for any x f(x)=g(x).

So, take an x. What do we have? Results about intervals. What might your first step be here?

How is this? Is this now a full proof?

Select any x then by IVT for any natural n [tex]\exists[tex]c_n in (x,x+1/n) such that
g(c_n)=f(c_n). we construct a sequence of such c_n values. Clearly the limit of c_n as n goes to infinity is x also the limit of f(c_n)= the limit of g(c_n). Hence by continuity g(x)=f(x).

Yes, this is good. Except, that sure it is clear that c_n-->x, but can you give a rigorous argument?

And also, why do you say "By IVT"?!?

To show c_n --> x could we just set epsilon=1/n? I don't know why I said by IVT.

That doesn't really make sense... to show that c_n-->x, you need to find, to every epsilon>0, an N>0 such that n>N ==> |c_n-x|<epsilon.

Yes, it suffices to find N for epsilon=1/m for all m>0 (by the archimedean property of R) but what is N for epsilon=1/m?

To prove that c_n-->x rigorously, you can either answer that question or use an argument involving the squeeze lemma (aka sandwich lemma)

How is this?

For a given episilon take N= 1 + the integer part of 1/epsilon.

Yup.