Timeline for Extensions of bounded uniformly continuous functions
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 12 at 19:07 | comment | added | TJP | @Jakobian Every bounded uniformly continuous pseudometric on a subspace of a uniform space extends to a uniformly continuous pseudometric on the whole space. See e.g. section 2.2 in "Uniform spaces and measures" by Jan Pachl (2013). | |
| Jul 17 at 14:14 | vote | accept | Jakobian | ||
| Jul 17 at 13:36 | comment | added | Jakobian | I see, so we are using that a uniformity generated by countable family of pseudometrics is generated by just one pseudometric (e.g. as in the proof that countable products of metrizable spaces are metrizable). This is definitely a trick worth remembering. Thank you | |
| Jul 17 at 9:45 | history | edited | KP Hart | CC BY-SA 4.0 |
one $A$ turned into an $S$
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| Jul 17 at 9:43 | comment | added | KP Hart | The $d_n$s are taken from the family of compatible pseudometrics on $X$, and $d$ is constructed on $X$ from the $d_n$. | |
| Jul 17 at 5:43 | history | edited | Jochen Wengenroth | CC BY-SA 4.0 |
Reduction explained
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| Jul 16 at 18:17 | comment | added | Jakobian | Could you explain why can we restrict to a pseudometric space $(X, d)$? I'm not sure if there has to exists an admissible pseudometric $d$ on $X$ such that for all $\varepsilon > 0$ there is $\delta > 0$ with $d(x, y)\leq \delta \implies |f(x)-f(y)|\leq \varepsilon$ for $x, y\in S$. Such pseudometric on $S$ surely exists, since $\Psi_f(x, y) = |f(x)-f(y)|$ is an example of one, but I am under impression that pseudometrics on subspace in general don't extend to the whole space (I'm using my notation). | |
| Jul 16 at 17:19 | history | answered | Jochen Wengenroth | CC BY-SA 4.0 |