Timeline for About the finished, $\aleph_0$...-compactness
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 16, 2019 at 13:01 | vote | accept | Dattier | ||
| Oct 16, 2019 at 12:38 | comment | added | Robert Furber | @Dattier The spread $s(E)$ is the supremum of the cardinalities of discrete subsets. So if $A \subseteq E$ has $|A| > s(E)$, it must have an accumulation point. | |
| Oct 16, 2019 at 12:36 | comment | added | Robert Furber | Under the definitions in the Handbook, what you're saying is "$E$ is separable iff $E$ has countable spread" (or $s(E) \leq \aleph_0$). In Theorem 8.1 (c) it is also shown that $s(E) = d(E) = L(E)$. | |
| Oct 16, 2019 at 12:34 | comment | added | Dattier | this theorem is knowing : ***SB theorem :***(E,d) is separable iff (for all A⊂E and card(A)>card(N) then A have an accumulation point) ? @Robert Furber | |
| Oct 16, 2019 at 12:31 | history | answered | Robert Furber | CC BY-SA 4.0 |