Timeline for Topological spaces with too many open sets
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 1, 2016 at 15:43 | vote | accept | alex alexeq | ||
| Aug 20, 2016 at 5:38 | comment | added | Forever Mozart | @JoelDavidHamkins you sunk my battleship :( | |
| Aug 19, 2016 at 16:09 | answer | added | Will Brian | timeline score: 4 | |
| Aug 19, 2016 at 15:42 | comment | added | მამუკა ჯიბლაძე | Such $X$ also cannot be metrizable. If you can extend $f(x):=\frac1{d(x,a)}$ with some $f(a)$, then in a small enough punctured neighborhood of $a$, $f(a)-\varepsilon<\frac1{d(x,a)}<f(a)+\varepsilon$, so that $f(a)$ is positive, and moreover $d(x,a)>\frac1{f(a)+\varepsilon}$ for all $x\ne a$ in a neighborhood of $a$. Whereas if $a$ is not isolated then in all of its punctured neighborhoods this distance will take values arbitrarily close to zero. | |
| Aug 19, 2016 at 14:26 | answer | added | Anonymous | timeline score: 2 | |
| Aug 19, 2016 at 11:45 | comment | added | Joel David Hamkins | No linear order topology can have the requested property, since for any $a$ we can let $f(x)=0$ for $x<a$ and $f(x)=1$ for $x\geq a$. Similarly, if deleting any point disconnects the space (with the point in the closure of two components), the same idea works. | |
| Aug 19, 2016 at 8:56 | comment | added | Forever Mozart | oh I see. Well, my idea would be to take a tree of long lines, so that each point is the $\omega_1$-point of some long line. | |
| Aug 19, 2016 at 8:55 | comment | added | alex alexeq | @ForeverMozart Thanks, but the property must be true for any $a \in X$ | |
| Aug 19, 2016 at 8:52 | comment | added | Forever Mozart | Yes. $X=\omega_1+1$ with order topology is such a space ($a=\omega_1$). If you want no isolated points, use the Long Line. | |
| Aug 19, 2016 at 8:48 | history | asked | alex alexeq | CC BY-SA 3.0 |