Uniform continuity

[Kavi Rama Murthy]

Let f be a map from R to R such that whenever d(A,B)=0 we have d(f(A),f(B))=0. Show that f is uniformly continuous.
Supposed to be a standard result but I cannot find a proof. Please help! [d(A,B) is the distance between A and B]
Kavi Rama Murthy

 

[pka]

Let f be a map from R to R such that whenever d(A,B)=0 we have d(f(A),f(B))=0. Show that f is uniformly continuous.

Suppose that \(\displaystyle f\) is not uniformly continuous on \(\displaystyle \mathbb{R} \).

The negation says \(\displaystyle \exists\alpha>0\) such that if \(\displaystyle \delta>0\) then \(\displaystyle \exists x~\&~y \) having the property that \(\displaystyle d(x,y)<\delta\to d(f(x),f(y))\ge\alpha.\).

For each \(\displaystyle n\in\mathbb{Z}^+ \) define \(\displaystyle x_n~\&~y_n \) such that \(\displaystyle d(x_n,y_n)<\frac{1}{n} \).

Let \(\displaystyle A=\{x_1,x_2,\cdots,x_n\cdots\}~\&~B=\{y_1,y_2 \cdots ,y_n,\cdots\} \)

What is \(\displaystyle d(A,B)=~? \)

 

[Kavi Rama Murthy]

Suppose that \(\displaystyle f\) is not uniformly continuous on \(\displaystyle \mathbb{R} \).

The negation says \(\displaystyle \exists\alpha>0\) such that if \(\displaystyle \delta>0\) then \(\displaystyle \exists x~\&~y \) having the property that \(\displaystyle d(x,y)<\delta\to d(f(x),f(y))\ge\alpha.\).

For each \(\displaystyle n\in\mathbb{Z}^+ \) define \(\displaystyle x_n~\&~y_n \) such that \(\displaystyle d(x_n,y_n)<\frac{1}{n} \).

Let \(\displaystyle A=\{x_1,x_2,\cdots,x_n\cdots\}~\&~B=\{y_1,y_2 \cdots ,y_n,\cdots\} \)

What is \(\displaystyle d(A,B)=~? \)

Sorry! Your argument does not answer the question. Of course, d(A,B)=0 and hence d(f(A),f(B))=0. But that does not lead to a contradiction since f(x_n) and f(y_m) may be close for some n and m but not necessarily with n=m.

 

[pka]

Sorry! Your argument does not answer the question.

I fully understand that. But as this site is usually used by students, I meant to give only basic direction on definitions. What I posted was never meant as a solution. We try not to do that here.

Actually, I doubt that the question is a theorem. But, I cannot find a counter-example.
I am surprise not have seen this before.
If you prove this one way or the other, I would really like to see the result.

 

[Kavi Rama Murthy]

Thank you for your interest in the problem The result is supposed to be true and several theorem are derived from this. In particular there are characterizations of metric spaces in which every bounded continuous function is uniformly continuous. I have see proofs of these in Americam Mathematical Monthly. They assume the validity of this basic result for functions from R to R and claim that this special case is well known. Seems to be a tricky problem.
Kavi Rama Murthy

I fully understand that. But as this site is usually used by students, I meant to give only basic direction on definitions. What I posted was never meant as a solution. We try not to do that here.

Actually, I doubt that the question is a theorem. But, I cannot find a counter-example.
I am surprise not have seen this before.
If you prove this one way or the other, I would really like to see the result.

 

[daon2]

pka is right on the money. I recall a theorem from my undrgraduate analysis class. A characterization for uniform continuity is: if \(\displaystyle \{x_n\},\{y_n\}\subseteq \mathbb{R}\) with \(\displaystyle d(x_n,y_n)\to 0\) then \(\displaystyle d\left(f(x_n),f(y_n)\right)\to 0\).

i'm late to the party, i see this doesn't work now. i'm willing to bet a modification of the sets might work