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!

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook