# Proof about uniformly continuous functions

1. Mar 8, 2009

### kivit

Let f be a uniformly continuous function on Q... Prove that there is a continuous function
g on R extending f (that is, g(x) = f(x), for all x∈Q

I think I am supposed to somehow use the denseness of Q and the continuity of a function to prove this, but I am not quite sure where I should start. Any help would be greatly appreciated.

2. Mar 8, 2009

### Focus

Pick an irrational t, what can you do with rationals to estimate it?

3. Mar 8, 2009

### Dick

First you have to define what g(x) is. Yes, use the denseness of Q. Can you do that?

4. Mar 9, 2009

### kivit

so would i be right to start out defining g as uniformly continuous on R?

then somehow using the denseness of Q to show that all the Qs are in R so g extends f?

... or am i not understanding what this means by extending?

5. Mar 9, 2009

### Dick

For every irrational t there are an infinite number of sequences q_i of rationals such that limit q_i->t. So you can define g(t)=limit f(q_i). Now you have to show that g(t) is independent of the sequence q_i. Use the continuity of f on the rationals. To show it's uniformly continuous, you can't define it to be. You have to show it is. Again use the uniform continuity of f.

6. Mar 10, 2009

### kivit

ok I think I can do it now. thanks :) I'll let you know how it goes in a little bit!