- #1

PsychonautQQ

- 784

- 10

## Homework Statement

Let f_n: [0,1]-->R be a uniformly convergent sequence of continuous functions. Let f: [0,1]-->R be the uniform limit of {f_n}. Let c be an element of R. Suppose that for all n there is some x_n in [0,1] such that f_n(x_n)=c. Prove that there exists an x in [0,1] such that f(x) = c

## Homework Equations

## The Attempt at a Solution

So the idea here is that there is always some x in [0,1] such that f_n(x)=c, and since f_n converges to f, it seems quite likely that there is an x such that f(x)=c. I need to formalize this proof, I'm not great at that, but let's go!

So a key element of this problem is that f_n is a UNIFORMLY convergence sequence and f is a uniform limit, therefore when searching for our epsilon (which we will call e), we can't have it depend on x.

let e > 0, then there is an N s.t. for all n>N |f_n(x)-f(x)|<e for all x in [0,1]. Let y_n be in [0,1] s.t. f_n(y_n)=c, then for all e>0 there is an N such that if n>N then |f_n(y_n) - f(y_n)| = |c - f(y_n)| < e ...

I'm confused, am i on the right track here?