# Characterize uniform continuity by sequences/net/filter

From Wikipedia

let $A$ be a subset of $\mathbb{R}^n$. A function $f : A → \mathbb{R}^m$ is uniformly continuous if and only if for every pair of sequences $x_n$ and $y_n$ such that $$\lim_{n\to\infty} |x_n-y_n|=0\,$$ we have $$\lim_{n\to\infty} |f(x_n)-f(y_n)|=0.\,$$

I was wondering if this can be generalized to $f : X → Y$ when $X$ is a metric space and $Y$ is $\mathbb{R}^m$ or even another metric space? If "if and only if" doesn't hold, does "if" or "only if" hold?

Are there generalizations when $X$ is a uniform space and $Y$ is $\mathbb{R}^m$ or even another uniform space? For example, by replacing sequence with net or filter, and distance with entourage?

Thanks and regards!

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## 2 Answers

Yes, indeed this can be generalized to metric spaces. Just replace the modulus by the corresponding metric in each case. The if and only if is still valid.

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The same result holds for maps between arbitrary metric spaces and the proof is word-for-word the same!

Theorem:

Let $(X,d)$ and $(Y, \rho)$ be metric spaces. The map $f: X \to Y$ is uniformly continuous if and only if for sequences $\{x_n\}$ and $\{y_n\}$ in $X$ such that $d(x_n,y_n) \to 0$, $\rho(f(x_n),f(y_n)) \to 0$

Proof:(Sketch)

One implication follows from the definition while for the other, suppose that the function is not uniformly continuous, construct sequence $\{x_n\}$ and $\{y_n\}$ such that $d(x_n,y_n) \to 0$ but $\rho(f(x_n),f(y_n))$ does not converge. (The construction becomes clear if you write the contrapositive!)

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Thanks! Can you define "arbitrary spaces"? –  Tim Feb 6 '12 at 16:38
That should have been arbitrary metric spaces! The very notion of convergence might not make sense on general topological spaces, as you will already know! In case you want me to write the proof, I am willing to do so. BTW, you ask questions on subject I am learning quite seriously! Good questions! –  user21436 Feb 6 '12 at 16:48
Thanks! Do you know about when dommain and/or codomain are uniform spaces? –  Tim Feb 6 '12 at 16:50
@Tim I don't know about these uniform spaces. I am sorry. I have just started learning general topology! –  user21436 Feb 6 '12 at 16:54
In a uniform space the definition of a uniformly continuous function $f: (X,\mathcal{U}) \rightarrow (Y, \mathcal{V})$ is: for every $V \in \mathcal{V}$ there exists $U \in \mathcal{U}$ such that $|x - y| < U$ implies $|f(x) - f(y)| < V$. What would your proposed filter reformulation be? –  Henno Brandsma Feb 6 '12 at 21:49