Timeline for Riesz representation for an infinite-dimensional space
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2013 at 23:12 | comment | added | Alexander Shamov | The set of compact subsets is naturally ordered by inclusion. To each compact $K$ we have to associate a space - namely, $C(K)$ - and for each pair $K_1 \subset K_2$ we need a map $C(K_2) \to C(K_1)$ - in this case it is the restriction map. | |
| Feb 4, 2013 at 22:12 | comment | added | Banach | I looked up projective limit, but I don't see why $C(X)$ is the projective limit of $C(K)$. Would you please indicate what are the coordinate spaces and the bonding maps in the inverse limit? | |
| Feb 4, 2013 at 20:43 | comment | added | Banach | According to my topology book, there is a restriction. The statement, "a function on X is continuous iff its restrictions on compact subsets are continuous" requires X to be compactly generated. However, in my case X is compactly generated. | |
| Feb 4, 2013 at 19:33 | comment | added | Alexander Shamov | @Banach: Of course not. A projective limit may be defined for any directed family, which may be neither countable nor linearly ordered. Intuitively, this means that a continuous function "is" a collection of functions on compact subsets that fit together nicely, and the topology on $C(X)$ is generated by taking these restrictions. For precise definitions, see, e.g., en.wikipedia.org/wiki/Inverse_limit and en.wikipedia.org/wiki/Initial_topology. | |
| Feb 4, 2013 at 9:53 | comment | added | Banach | What does it mean that $C(X)$ is the projective limit of $C(K)$? Does this require X to be a countable union of compact sets? | |
| Feb 3, 2013 at 23:52 | history | answered | Alexander Shamov | CC BY-SA 3.0 |