Uniform continuity, Cauchy sequences

Aug 2009
24
8
If D ⊂ R^p, prove that function F : D → R^q is uniformly continuous on
D if and only if {F(x_n)} is a Cauchy sequence in R^q whenever {x_n} is a Cauchy sequence in D

I proved it =>, but now im having problems with <=. is this statement even true? or does the set D must be bounded?
 

Bruno J.

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Jun 2009
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It's false : for instance \(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R} : x \mapsto x^2\) has the property of taking Cauchy sequences to Cauchy sequences, but is not uniformly continuous on \(\displaystyle \mathbb{R}\). The statement holds if \(\displaystyle D\) is compact and I believe compact sets are the most general sets for which it holds; I'm too tired to attempt justifying this assertion now, maybe tomorrow!
 

Drexel28

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Nov 2009
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It's false : for instance \(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R} : x \mapsto x^2\) has the property of taking Cauchy sequences to Cauchy sequences, but is not uniformly continuous on \(\displaystyle \mathbb{R}\). The statement holds if \(\displaystyle D\) is compact and I believe compact sets are the most general sets for which it holds; I'm too tired to attempt justifying this assertion now, maybe tomorrow!
I too am too tired to check, but while compactness is sufficient we may possibly only need totally bounded. This has no real backing as of now but my own sleepy thoughts.
 

Bruno J.

MHF Hall of Honor
Jun 2009
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498
Canada
I too am too tired to check, but while compactness is sufficient we may possibly only need totally bounded. This has no real backing as of now but my own sleepy thoughts.
You're right! Here's my proof.

Suppose \(\displaystyle D\) is bounded and \(\displaystyle f\) takes Cauchy sequences to Cauchy sequences. I claim that \(\displaystyle f\) can be extended to a continuous function on \(\displaystyle \overline D\), which is compact; then \(\displaystyle f\) is uniformly continous on \(\displaystyle \overline D\) and thus certainly on \(\displaystyle D\) also.

It suffices to show that if \(\displaystyle (x_n), (y_n)\) are two Cauchy sequences with the same limit \(\displaystyle d \in \overline D\), then \(\displaystyle (f(x_n)),(f(y_n))\) have the same limit which we may then call \(\displaystyle f(d)\). Consider the sequence \(\displaystyle (x_1, y_1, x_2, y_2,\dots)\), which converges to \(\displaystyle d\) also; by assumption, it is mapped by \(\displaystyle f\) to a convergent sequence in \(\displaystyle \mathbb{R}^q\), which implies the subsequences \(\displaystyle (f(x_n)),(f(y_n))\) have the same limit!
 

Drexel28

MHF Hall of Honor
Nov 2009
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Berkeley, California
You're right! Here's my proof.

Suppose \(\displaystyle D\) is bounded and \(\displaystyle f\) takes Cauchy sequences to Cauchy sequences. I claim that \(\displaystyle f\) can be extended to a continuous function on \(\displaystyle \overline D\), which is compact; then \(\displaystyle f\) is uniformly continous on \(\displaystyle \overline D\) and thus certainly on \(\displaystyle D\) also.

It suffices to show that if \(\displaystyle (x_n), (y_n)\) are two Cauchy sequences with the same limit \(\displaystyle d \in \overline D\), then \(\displaystyle (f(x_n)),(f(y_n))\) have the same limit which we may then call \(\displaystyle f(d)\). Consider the sequence \(\displaystyle (x_1, y_1, x_2, y_2,\dots)\), which converges to \(\displaystyle d\) also; by assumption, it is mapped by \(\displaystyle f\) to a convergent sequence in \(\displaystyle \mathbb{R}^q\), which implies the subsequences \(\displaystyle (f(x_n)),(f(y_n))\) have the same limit!
I think we can prove something more general. If \(\displaystyle f:E\to\mathcal{N}\) preserves Cauchy sequences, with \(\displaystyle \mathcal{M},\mathcal{N}\) metric spaces with \(\displaystyle \mathcal{M}\) complete. and \(\displaystyle E\subseteq\mathcal{M}\), then a) it is continuous and we may extend it to a continuous function \(\displaystyle f:\mathcal{M}\to\mathcal{N}\).

I know this because I had to prove it once (I'll find the proof) tomorrow, and on second thought we might need \(\displaystyle E\) dense in \(\displaystyle \mathcal{M}\)