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

Let me check on that.

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

Dear Vladimir Kadets,

Thank you! Thus, no doubts now:)

Best regards, Joachim