Timeline for Metrization of spaces of functions
Current License: CC BY-SA 3.0
9 events
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| Sep 17, 2013 at 10:41 | comment | added | Pietro Majer | I think one can repeat to some extent the construction of the compact-open topology in $C(M,N)$ by means of an ideal of sets $\mathcal{I}$ on $M$ (instead of the family of compacts of $X$). The metrizability of $N$ and the countable cofinality of $\mathcal{I}$ should be again the condition for metrizability; one gets the distance of uniform convergence on elements of $\mathcal{I}$. | |
| Sep 16, 2013 at 22:22 | answer | added | Włodzimierz Holsztyński | timeline score: 1 | |
| Sep 16, 2013 at 22:05 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tag 'topology' and other tag
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| Sep 16, 2013 at 20:30 | vote | accept | Markovjan | ||
| Sep 16, 2013 at 20:27 | answer | added | Pietro Majer | timeline score: 11 | |
| Sep 16, 2013 at 20:25 | comment | added | Markovjan | Im trying to obtain the less restrictive conditions on the topological properties on $M$, $N$, (Combinations of second- countable, lindelöf, hausdorff, paracompact or some othe properties) such that $C(M,N)$ is metrizable with some dynamically relevant topology in the sense that the stability notion of dynamical systems (en.wikipedia.org/wiki/Topological_conjugacy) is not vacuous. | |
| Sep 16, 2013 at 20:09 | comment | added | Igor Khavkine | For the metrization question to make sense, you first need to start with some topology on $C(M,N)$ (e.g., compact-open). Some metrization conditions are well known (en.wikipedia.org/wiki/Metrization_theorem, en.wikipedia.org/wiki/…) so if $C(M,N)$ fits them, it is metrizable. But perhaps this is not the kind of answer you are looking for... In that case, what is the motivation behind the question? | |
| Sep 16, 2013 at 19:34 | review | First posts | |||
| Sep 16, 2013 at 20:12 | |||||
| Sep 16, 2013 at 19:15 | history | asked | Markovjan | CC BY-SA 3.0 |