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A family of pseudometrics defined on a set gives rise to a uniform structure on that set. Moreover (up to uniform equivalence, anyway) every uniform structure arises this way. Let $A$ and $B$ be families of pseudometrics defined on sets $X$ and $Y$ respectively. There ought to be a way to characterize the uniform continuity of maps $X$ to $Y$ directly in terms of $A$ and $B$ and I thought I knew the correct way to do this but, when I scanned through the relevant sections in Dugundji's Topology to confirm my suspicions, I came across:

"$f: X \to Y$ is uniformly continuous if for each $\beta \in B$ and each $\epsilon > 0$, there exists an $\alpha \in A$ and a $\delta > 0$ such that if $\alpha(x,x') < \delta$, then $\beta(f(x),f(x')) < \epsilon$."

This seems stronger than uniform continuity to me. When I tried to work out the right condition I found that one might, given an $\epsilon > 0$ and a $\beta \in B$, only be able to find a $\delta>0$ and finitely many pseudometrics $\alpha_1,\ldots,\alpha_n \in A$ such that $(\alpha_i(x,x') < \delta \ \ \forall i) \Rightarrow \beta(f(x),f(x')) < \epsilon$.


Edit: I don't think there's much more one can say in response to this question. If $\mathscr{D}$ is a family of pseudometrics on $X$, one can safely replace $\mathscr{D}$ with the family $\mathscr{D}^+$ consisting of all pseudometrics of the form $(x,y) \mapsto \max_{d \in \mathscr{F}} d(x,y)$ where $\mathscr{F}$ is some finite subset of $\mathscr{D}$ without inducing a finer uniform structure on $X$. As remarked by Theo Buehler in the comments, if we assume, with no loss of generality so far as the uniform structure on $X$ is concerned, that my $A = A^+$ then Dugundji's definition of uniform continuity is equivalent to the clumsier one. In summary, there is a small inaccuracy in Dugundji which can be easily corrected by the addition of a single superscript $+$ to make explicit an assumption that the author (rather harmlessly I might add) probably considered implicit. I would call the matter settled. If anyone disagrees, feel free to say so, but you will do so too late to save my hapless copy of Dugundji - which has already been permanently defaced!

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For example: I thought the $L^\infty$ norm gave the same uniform structure on $\mathbb{R}^2$ as the family of seminorms $\{(x,y) \mapsto |x|,(x,y) \mapsto |y|\}$. If so, the identity map $\mathbb{R}^2 \to \mathbb{R}^2$ should be uniformly bicontinuous in this situation... – Mike F Mar 23 '11 at 3:01
Is one of those "d"s supposed to be d' ?? – The Chaz 2.0 Mar 23 '11 at 4:05
@The Chaz: Sorry, I posted too quickly I guess. I think I've ironed out the wrinkles now. – Mike F Mar 23 '11 at 4:48
Can't you just replace the set $A$ in your condition by the set $A_f$ of finite subsets of $A$ and take $\alpha_I = \max_{i \in I} \alpha_i$ for $I \in A_f$ to get Dugundji's condition? – t.b. Mar 23 '11 at 6:32
You can do that safely as the example in your first comment shows. On the other hand, I prefer to use a pencil in such situations anyway. – t.b. Mar 23 '11 at 7:42
up vote 1 down vote accepted

This was resolved in the comments

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