Timeline for Between an open set and its closed subset [closed]
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2021 at 21:03 | comment | added | მამუკა ჯიბლაძე | @IosifPinelis Oh I see now, sorry! By some reason I was reasoning as if $F$ is open rather than closed... | |
| Oct 11, 2021 at 19:53 | comment | added | Iosif Pinelis | @მამუკაჯიბლაძე : Yes, in a regular space $X$, for any nonempty open $V$ there is a nonempty open $U$ with $\overline U\subseteq V$. But this is not enough for the normality of $X$, even if $X$ is Hausdorff. And the posted problem is equivalent to that of proving that the topological space $\mathbb R$ is normal. | |
| Oct 11, 2021 at 19:07 | comment | added | მამუკა ჯიბლაძე | @IosifPinelis In a regular space, for any nonempty open $V$ there is a nonempty open $U$ with $\overline U\subseteq V$, right? | |
| Oct 10, 2021 at 21:46 | comment | added | Iosif Pinelis | @მამუკაჯიბლაძე : I think the regularity is not enough. There are regular Hausdorff spaces that are not normal math.stackexchange.com/a/3874422/96609 | |
| Oct 10, 2021 at 10:42 | history | left closed in review |
Jeremy Rickard Alex M. Daniele Tampieri |
Original close reason(s) were not resolved | |
| Oct 10, 2021 at 6:50 | comment | added | მამუკა ჯიბლაძე | A topological space is regular if and only if every open $V$ is the union of those opens $U\subseteq V$ satisfying $\overline U\subseteq V$. Then use that $\mathbb R\setminus F$ is regular. | |
| Oct 10, 2021 at 4:55 | vote | accept | Analyst_311419 | ||
| Oct 10, 2021 at 4:17 | review | Reopen votes | |||
| Oct 10, 2021 at 10:42 | |||||
| Oct 10, 2021 at 2:03 | history | closed |
LSpice user44191 R W Moishe Kohan Andreas Blass |
Not suitable for this site | |
| Oct 10, 2021 at 1:46 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Oct 9, 2021 at 22:36 | review | Close votes | |||
| Oct 10, 2021 at 2:05 | |||||
| Oct 9, 2021 at 22:17 | history | edited | LSpice | CC BY-SA 4.0 |
`\cup` -> `\bigcup`
|
| Oct 9, 2021 at 22:16 | comment | added | LSpice | You choose $r_x$, use it to define $U$, then define $r_y$ depending on $U$—that's circular. (The claim that $U \subsetneq V$ also seems not to have been demonstrated.) This is not research level. | |
| Oct 9, 2021 at 22:12 | history | asked | Analyst_311419 | CC BY-SA 4.0 |