Timeline for a characterisation of proper maps via ultrafilters
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2016 at 11:47 | vote | accept | user97621 | ||
| Sep 16, 2016 at 14:10 | comment | added | Taras Banakh | In my example the embedding $A\subset B$ is not ultrafilter-like. So, sorry! | |
| Sep 16, 2016 at 14:04 | answer | added | Taras Banakh | timeline score: 1 | |
| Sep 15, 2016 at 10:44 | comment | added | user97621 | I am a bit lost in notation: above it said $X=\beta\omega$... Let me try to start from scratch. We have two free ultrafilters $a,b \in\beta\omega$, $a\neq b$. the proper map is $\beta\omega\rightarrow \beta\omega/\{a,b\}$. the ultrafilter-like map is $\beta\omega\setminus\{a,b\}\rightarrow \beta\omega$. Is this correct? | |
| Sep 15, 2016 at 10:38 | comment | added | user97621 | I am a bit lost in notation: above it said $X=\beta\omega$... Let me try to start from scratch. We have two free ultrafilters $a,b \in\beta\omega$, $a\neq b$. | |
| Sep 15, 2016 at 6:20 | comment | added | Taras Banakh | The topology on the domain $X$ is the subspace topology inherited from $\beta\omega$. The qotient map $q:\beta\omega\to Y$ restricted to $X=\beta\omega\setminus\{a,b\}$ is a topological embedding of $X$ into $Y=B$ (because of the properness of $q$. By the way, in my first comment I made a mistake defining $q$ as the qutient map from $X$. It should be the quotient map from $\beta\omega$. | |
| Sep 13, 2016 at 6:19 | comment | added | user97621 | I am sorry for delay. what is the topology on the domain of the ultrafilter-like map? it is not the induced topology? I require that the topology on subset A is induced from B | |
| Aug 31, 2016 at 20:26 | comment | added | Taras Banakh | And what is wrong with $q(a),q(b)\in Y$? | |
| Aug 31, 2016 at 15:22 | comment | added | user97621 | so is it $\beta\omega\rightarrow \beta\omega / \{a,b\}$ and $\beta\omega\setminus \{q(a),q(b)\} \rightarrow \beta\omega/ \{a,b\}$ ? (there is another misprint in the first map, nothing follows $\beta$) I do not see it, as $q(a),q(b)\in Y$ are elements of the quotient.. | |
| Aug 31, 2016 at 15:08 | comment | added | Taras Banakh | $\beta\omega\to\beta/\{a,b\}$ is the proper map. $\beta B\setminus\{q(a),q(b)\}$ should be $\beta\omega\setminus\{q(a),q(b)\}$. Sorry for this misprint. | |
| Aug 31, 2016 at 6:50 | comment | added | user97621 | I do not understand your notation: which is the ultralifter-like map and which is the proper map ? $\beta\omega\rightarrow\beta\omega/\{a,b\}$ is the proper map and $\beta B\setminus\{q(a),q(b)\} \rightarrow \beta\omega/\{a,b\}$ is the ultrafilter-like map ? but the latter is not a subset, is it? | |
| Aug 30, 2016 at 13:29 | comment | added | Taras Banakh | How about the following (possible) counterexample: Let $a,b$ be two distinct free ultrafilters in $X=\beta\omega$, $Y=X/\{a,b\}$ be the quotient space and $q:X\to Y$ be the quotient map. Let $B=Y$ and $A=\beta B\setminus \{q(a),q(b)\}$ and $f:A\to X\setminus\{a,b\}$, $g:B\to Y$ be the identity maps. It seems that no map $h$ with the required properties exists. Is it Ok? | |
| Aug 24, 2016 at 8:33 | review | First posts | |||
| Aug 24, 2016 at 8:45 | |||||
| Aug 24, 2016 at 8:28 | history | asked | user97621 | CC BY-SA 3.0 |