Timeline for A criterion for metrizable topological spaces [duplicate]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2018 at 15:15 | comment | added | Pietro Majer | Also, inductive limits of metric spaces in general are not metrizable. A typical obstruction is given by Baire theorem. | |
| Apr 26, 2018 at 12:27 | comment | added | Todd Trimble | @bof Yes, which means the answer to OP's question is false even if we restrict to increasing chains of metrizable subspaces (e.g., the integers with the cofinite topology). | |
| Apr 26, 2018 at 12:21 | history | closed | Todd Trimble | Duplicate of Are countable unions of metrizable spaces metrizable too? | |
| Apr 26, 2018 at 9:17 | comment | added | bof | Isn't every countable $\text{T}_1$-space the union of a sequence of metrizable spaces? | |
| Apr 26, 2018 at 8:48 | comment | added | ABB | @Corbennick Actually I faced a particular type of topological spaces and would like to know whether they are well-known or not: Let $X$ be a topological space. Let us say $X$ has property P if there exists a metric $d$ on $X$ and a sequence of spaces $X_n\subseteq X$ such that all $X_n$ are $d$-metrizable and $X=\bigcup X_n$. | |
| Apr 26, 2018 at 8:35 | comment | added | ABB | @ Corbennick Nice example. | |
| Apr 26, 2018 at 8:30 | comment | added | ABB | @ Evgeny Kuznetsov I mean $X_n$ is just contained in $X$ for every $n$. | |
| Apr 26, 2018 at 8:26 | comment | added | user1688 | The line with double origin is a union of two metrizable spaces, but is not metrizable itself, as it is not Hausdorff. | |
| Apr 26, 2018 at 8:22 | comment | added | Evgeny Kuznetsov | Please explain, what does it mean a sequence of spaces? | |
| Apr 26, 2018 at 8:19 | history | asked | ABB | CC BY-SA 3.0 |