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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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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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    @HennoBrandsma: Thanks, but I don't know. Were you suggesting it is not possible to have an (not necessarily) equivalent concept in terms of filter or net? By the way, I think the two distances should be replaced with pairs of points.2012-02-06
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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.