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Let $X$ be a uniform space and $F(X)$ the vector space of all uniformly continuous real-valued functions over $X$. It is possible to express every bounded uniform semimetric $d$ on $X$ as $d(x,y) = d_B(x,y) := \sup_{f\in B} |f(x)-f(y)|$ with a suitable pointwise bounded and uniformly equicontinuous $B\subseteq F(X)$, choose e.g. $B = \{\,f_{\hat{x}}:X\rightarrow\mathbb{R}\,\,|\,\,\hat{x}\in X\,\}$ with $f_{\hat{x}}(x) := d(\hat{x},x)$.

In general, there should be more than only one uniform structure on $X$ that give rise to the same $F(X)$. The weakest one is clearly the weak uniform structure induced by $F(X)$ on $X$, i.e. the uniform structure given by the uniform semimetrics $d_{\{f\}}$ for all $f\in F(X)$. Is there also a strongest one (analogous to the Mackey-topology in the theory of locally convex spaces)?

Moreover, I would conjecture that (unlike for locally convex spaces) $X$ is complete if and only if it is complete under the weak uniform structure, because (unlike for locally convex spaces) the induced topologies are the same, namely the weak topology induced on $X$ by $F(X)$. Is this true?

I would also be very happy if someone could point out a good source for learning about uniform spaces that goes beyond a mere appendix to the theory of topological spaces.

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    $\begingroup$ Typo/mis-spelling in title $\endgroup$ – Yemon Choi Oct 12 '15 at 16:17
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The answer is no for your second question if we assume sufficiently strong large cardinal hypotheses. Let $X$ be a set of measurable cardinality and then give $X$ the discrete uniformity. Then $X$ is complete in this uniformity. However, if we give $X$ the weak uniform structure, then $X$ is not complete in the weak uniform structure. In fact, the completion of $X$ is the Hewitt realcompactification $\upsilon X$ of $X$ which is the set of all $\sigma$-complete ultrafilters on $X$.

As for learning about uniform spaces, the books on uniform spaces by I.M James and John Isbell are the two books that deal exclusively with uniform spaces. The general topology text by Willard also has a large chapter on uniform spaces which might be helpful if you are not already familiar with that text.

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  • $\begingroup$ Thanks... This is not really the easy counterexample I have hoped for, but then there is no point in trying to prove this. $\endgroup$ – Matthias Oct 14 '15 at 20:23
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A great reference for Uniform Spaces is Isbell's book:

http://www.amazon.com/Uniform-Spaces-Mathematical-Surveys-Monographs/dp/0821815121

There are also two MO questions that provided references:

A good place to read about uniform spaces

Category of Uniform spaces

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  • $\begingroup$ I didn't have time to try and answer your questions but will have a look tomorrow if this is still unanswered $\endgroup$ – David White Oct 12 '15 at 19:38
  • $\begingroup$ Okay, I found a free moment. Isbell's book discusses preuniformities on page 5, and proves they form a complete lattice. So there's a strongest and a weakest, and this might help with your question 1. As for your second question, Joseph shows the answer is no in general, but I.M. James has an entire chapter on the topic (Chapter 12), so perhaps you can read what is known there. $\endgroup$ – David White Oct 13 '15 at 23:37
  • $\begingroup$ Thanks, but from the complete lattice property I only get that the pre-uniformities that give rise to the same unif. cont. real functions have a supremum, how do I see that this is in fact a maximum? $\endgroup$ – Matthias Oct 14 '15 at 20:14
  • $\begingroup$ This supremum should have the union of all the collections of entourages as a subbase, by taking finite intersections there might arise new entourages that in the end allow for more unif. cont. functions. But maybe I can show that this does not happen... $\endgroup$ – Matthias Oct 14 '15 at 20:20
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Another counterexample for the conjecture in paragraph 3: The closed unit ball $X$ in an infinite-dimensional Banach space with the metric of its norm is complete but the weak uniformity from $F(X)$ is not.

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