# Uniform Continuity (and Irrationals)

1. Oct 11, 2004

### zolit

I am a little shaky with the concept of proving uniform continuity vs regular continuity. Is the difference when proving through epsilon-delta definition just that your delta can not depend on "a" (thus be defined in terms of "a") (when |x-a|<delta) for uniform continuity?

Also to the more specific questions: if you have a function that is uniformly continuous for all rational numbers how can you prove that an extension of that function is uniformly continuous on all real numbers. I was thinking that from density of rationals and irrationals I can say that a rational number is necessarily flanked by two irrationals. Thus if we treat the rational number as an open interval between those two irrationals (interval which happens to contain only one point) then necessarily the function on the closed interval including them is also uniformly continuous. But I feel like that doesn't prove it for ALL irrationals - since technically between two rationals there's an infinite number of irrationals right? Am i totally off track here?

2. Oct 11, 2004

### Hurkyl

Staff Emeritus
Right, more or less. Asserting continuity on the domain U goes, the statement is:
For all x in U, for all e > 0, there is a d > 0, for all y in U: ...
whereas for uniform continuity they go
For all e > 0, there is a d > 0, for all x in U, for all y in U: ...

Where the ... is, of course, 0 < |x - y| < d ==> |f(x) - f(y)| < e

First off, I'm assuming you mean to continuously extend your rational function to make a real function. First you have to prove you can do that.

Also, keep in mind that between any two distinct irrational numbers, there are infinitely many rational numbers. However, you're close to the right idea; you're trying to make use of some sort of concept that irrational numbers are "close" to rational numbers. What you're trying to say is that:

For any irrational number p, and for any d > 0, there is a rational number q such that |p - q| < d.

That should help prove what you want to prove.