Timeline for Continuous functions as uniformly continuous function
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2013 at 15:26 | comment | added | Pietro Majer | Excellent. (Sorry for the variant of the first part, I had the idea while far from a computer and recalled wrongly) | |
| Sep 10, 2013 at 12:46 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
generalize
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| Sep 10, 2013 at 12:12 | comment | added | Emil Jeřábek | The first part of the proof already applies to arbitrary topologies as is (that was intentional). Thanks for the second part; I was trying in vain to show that any such topology has to coincide with the Euclidean topology, I did not realize that one could construct a non-uniform continuous counterexample regardless of that. | |
| Sep 10, 2013 at 8:43 | comment | added | Pietro Majer | The first part of your proof can also be done for general topologies, that is, $C((\mathbb{R},\tau),(\mathbb{R},\tau))=UC(\mathbb{R},\mathbb{R})$ implies $\tau$ is finer than the Euclidean topology. As before such $\tau$ is not the indiscrete topology and it is translation invariant. So there is a non-empty $\tau$-open set $G$ not containing $0$. Then there is also a non-empty bounded $\tau$-open set, e.g. of the form $f^{-1}(G)$ with $f$ continuous with compact support. Since $\tau$ is also homotety invariant, this implies that $\tau$ is finer than the Euclidean topology. | |
| Sep 10, 2013 at 8:33 | comment | added | Pietro Majer | A slight variant of the second part of above argument: since the $d$-topology is finer than the Euclidean, the family of intervals $I_n:=[n,n+1]$ for $n\in\mathbb{N}$ is a locally finite closed cover of $\mathbb{R}$ in the $d$-topology. Let $f$ be $x\mapsto x^2$. On each $I_n$, $f$ coincides with some uniformly continuous function, which is $d$-continuous. So $f_{|I_n}$ is also $d$-continuous, hence $f$ is $d$-continuous, hence $U$-continuous, contradiction. This way the argument works more generally for any topology on $\mathbb{R}$, even non-metric. | |
| Sep 9, 2013 at 18:31 | vote | accept | CommunityBot | ||
| Sep 9, 2013 at 15:45 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |