Expert: Paul Klarreich - 9/19/2007

Question
Prove ,If function 'f' is continuous on a closed set 'D',then 'f' is uniformly continuous on 'D'.

Answer
Questioner:   prem
Category:  Calculus
Private:  No
 
Subject:  continuity
Question:  Prove ,If function 'f' is continuous on a closed set 'D',then 'f' is uniformly continuous on 'D'.
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Hi, Prem,

This is really an oldie.  (For me, that is.)  As I recall, definitions are:

[I can't make epsilons and deltas; I have to use  e and d.]

f(x) is continuous at  x = x0  if given  e > 0, there exists  d > 0  such that
| f(x) - f(x0) | < e  whenever  | x - x0 | < d.

Ordinarily, the choice of d would depend both on e and on x0.

f(x) is UNIFORMLY continuous on D  if given  e > 0, there exists  d > 0  such that for all a in D, | f(x) - f(x0) | < e  whenever  | x - x0 | < d.

That is, the choice of d would depend on e, but not on x0 -- one size d fits all x0's.

Alas, one of my children seems to have taken my old copy of Buck and my Taylor, and neither is available for consultation right now, so I shall have to wing it.

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I recall that this depends on the Heine-Borel theorem, which says (and whose proof you can certainly look up in your advanced calculus - analysis text):

A set is closed and bounded if and only if it is compact.  

Definition: A set D is compact if given any collection of open sets X that 'covers' D, some finite subcollection of those sets will cover D.  

i.e. If you have an infinite collection of open sets whose union is D, you never need an infinite number of them.

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Now you have to prove:  If  f is continuous on a closed AND BOUNDED set D,then 'f' is uniformly continuous on D.

-- see note later.

Given your f(x), continuous on D, choose some e > 0.  Now for every  a in D, we can find  d > 0 such that:

| f(x) - f(a) | < e  whenever  | x - a | < d.

Now what is | x - a | < d ?  This is an open set; it is the set:

I(a) =  a - d < x < a + d

Since each 'a' certainly belongs to its corresponding I(a), the union of the I(a)'s must cover D.

Therefore, by the H-B theorem, some finite collection of these I(a)'s covers D.  In that case, we have a finite set of d's and we can pick the smallest of them as our  'd' that will work for every value of a.

There may be some details to work out but I think you can handle that.  If you still have trouble with it, let me know.  (By that time one of my children may be available.)
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Note: Yes, you need the 'and bounded' property; f(x) = x^2 is continuous on [0,infinity], which is a closed set, but not uniformly continuous there.