Dear all,
we know, owing to a result appearing in the book of R. M. Dudley (Real analysis and Probability - Theorem 2.8.2), that for any separable metric space (S, d), there is a metric e on S, defining the same topology as d, such that (S, e) is totally bounded.
I would like to know if a uniform continuous function f for the metric d will remain uniform continuous for the constructed metric e.
Regards,
Nathalie
Consider some unbounded uniformly continuous function f on an (unbounded) separable metric space (S,d). If f remained uniformly continuous on (S,e) it would extend to the completion of (S,e) which is compact, thus would be bounded.
Consider some unbounded uniformly continuous function f on an (unbounded) separable metric space (S,d). If f remained uniformly continuous on (S,e) it would extend to the completion of (S,e) which is compact, thus would be bounded.
Dear Michal,
thank you for your clear and simple counter-example.
Regards
I think the following example gives a direct answer to Nathalie's question, and completes in a way the interesting Michal's sketch:
For S=R:=(-\infty, \infty), d(x, y):= |x-y|, e(x,y) := | tanh x - tanh y|, f(x) := x for x \in S, by the Lagrange formula and by monotonicity of cosh, one obtains:
if x,y> M >0 and |x - y| >M, then |f(x) - f(y)| > cosh^2(M) e(x,y),
which proves (by arbitrariness of M), that
-- f is NOT uniformly continuous map from (S,e) onto (R,d),
-- f IS uniformly continuous as a map from (S,d) onto (R,d).
In the above example, the completion of (S,e) is compact. BUT, is this always the case? In this respect I share the doubts expressed by Thawatchai. As an example one can take for S the closed unit ball in C[0,1], which is complete with respect to the metric determined by the sup_norm, which in turn is totally bounded, but the space is not compact. Is in this any missed point?
Regards, Joachim
Concerning the doubts of Joachim Domsta. There is a misunderstanding of the terminology. The words "totally bounded" mean that for every epsilon there is a finite epsilon-net (pre-compactness in other terminology). In this sense the ball B of C[0,1] is not totally bounded. See
https://en.wikipedia.org/wiki/Totally_bounded_space#Definition_for_a_metric_space
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