Uniform Continuity (and Irrationals)

[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?

 

[Hurkyl]

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?

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

Also to the more specific questions: ...

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.

 

[M98Ranger]

The concept of uniform continuity can be a bit tricky to grasp at first, but let's break it down step by step.

First, let's review the definition of continuity: A function f is continuous at a point a if for any ε>0, there exists a δ>0 such that |x-a|<δ implies |f(x)-f(a)|<ε. This means that for any small change in x (represented by δ), there is a corresponding small change in f(x) (represented by ε).

Now, for uniform continuity, the definition is slightly different. A function f is uniformly continuous on a set S if for any ε>0, there exists a δ>0 such that for all x,y∈S, |x-y|<δ implies |f(x)-f(y)|<ε. Notice that for uniform continuity, the delta must work for all points in the set S, not just a specific point a. This is the key difference between uniform continuity and regular continuity.

To answer your first question, yes, the difference when proving through the epsilon-delta definition is that for uniform continuity, delta cannot depend on a specific point a, but must work for all points in the set S.

Now, onto your specific question about proving uniform continuity for an extension of a function. Your approach is on the right track, but you are correct in thinking that it doesn't prove it for all irrationals. Remember, the definition of uniform continuity requires that the delta works for all points in the set S. In this case, the set S is the set of all real numbers, including both rationals and irrationals. So, while your approach works for the rationals, it doesn't necessarily hold for all irrationals.

One way to prove uniform continuity for an extension of a function is to use the fact that the function is already uniformly continuous on a subset of the reals (in this case, the rationals). This allows you to choose a delta that works for all points in that subset, and then you can extend that delta to the entire set of real numbers. This is known as the "epsilon-delta game" and is a common approach in proving uniform continuity for extensions of functions.

In summary, the key difference between uniform continuity and regular continuity is that for uniform continuity, the delta must work for all points in the set, not just a specific point. And for proving uniform continuity