Timeline for Pointwise convergence and disjoint sequences in $C(K)$
Current License: CC BY-SA 4.0
17 events
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| Dec 16, 2023 at 4:40 | vote | accept | erz | ||
| Sep 23, 2023 at 20:21 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 23, 2023 at 14:53 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 21, 2023 at 11:26 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 21, 2023 at 11:01 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 21, 2023 at 10:55 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 21, 2023 at 10:49 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 20, 2023 at 16:41 | history | edited | user495577 | CC BY-SA 4.0 |
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| Sep 18, 2023 at 18:04 | comment | added | user495577 | Pelczynski and Semadeni in Spaces of Continuous Functions (III) showed that for a compact, Hausdorff space $K$, $K$ is scattered (which they call dispersed, which is equivalent to having no perfect, non-empty subset) if and only if $C(K)$ contains no isometric copy of $C([0,1])$. Metrizability is not needed here. Therefore if $K$ is not scattered, $K$ contains a an isometric copy of $C([0,1])$, and therefore an isometric copy of $\ell_2$. If $(f_n)_{n=1}^\infty\subset C(K)$ is equivalent to the canonical $\ell_2$ basis, it is pointwise null and has no subsequence which is almost disjoint. | |
| Sep 18, 2023 at 17:00 | history | undeleted | user495577 | ||
| Sep 18, 2023 at 16:32 | history | deleted | user495577 | via Vote | |
| Sep 18, 2023 at 16:07 | review | First answers | |||
| Sep 18, 2023 at 16:25 | |||||
| S Sep 18, 2023 at 16:06 | review | First answers | |||
| Sep 18, 2023 at 16:07 | |||||
| S Sep 18, 2023 at 16:06 | history | edited | user495577 | CC BY-SA 4.0 |
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| S Sep 18, 2023 at 15:50 | review | First answers | |||
| Sep 18, 2023 at 15:53 | |||||
| S Sep 18, 2023 at 15:50 | history | answered | user495577 | CC BY-SA 4.0 |