Timeline for $T_2$ topologies that are "as disjoint as possible"
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2016 at 6:45 | comment | added | Dominic van der Zypen | @LajosSoukup - You're right, sorry I was mistaken | |
| Mar 29, 2016 at 18:45 | comment | added | Will Brian | @DominicvanderZypen: No worries -- I agree that this answer deserves to be accepted. | |
| Mar 25, 2016 at 11:08 | comment | added | Lajos Soukup | Dominic, I can not see that connectedness is important here because I think that using my argument one can prove the following statement: If $X=(X,\rho)$ s.t. $$z(X)< |X|=|\rho|=min\{|U|:U\in \rho, U\ne \emptyset \}\le 2^{2^\omega}$$ then there is a topology $\tau$ on $X$ s.t $\rho$ and $\tau $ intersect in only the co-finite sets. | |
| Mar 24, 2016 at 7:50 | vote | accept | Dominic van der Zypen | ||
| Mar 24, 2016 at 7:50 | comment | added | Dominic van der Zypen | I realise that having asked 2 questions that each get an answer in a different post puts me in the difficult position to select the "canonical" accepted answer. I decided that since Lajos found the paper that Will's answer is based on, and since Lajos put in a lot of work to prove the result in this post, he deserves to get his answer the accepted one, so I changed acceptance. Will - I hope this doesn't annoy you too much! | |
| Mar 24, 2016 at 7:45 | comment | added | Dominic van der Zypen | Wonderful, thanks a lot Lajos...! Intuitively speaking, could it be that since the Euclidean topology $\varepsilon$ is "very connected" (in a quite vague sense of the word), a $T_2$-topology that is "as disjoint as it gets" from $\varepsilon$ must be "very disconnected" (such as the one you constructed? But maybe my intuition is just rubbish :-) | |
| Mar 23, 2016 at 17:28 | comment | added | Will Brian | Very nice! And this argument seems to apply to lots of other familiar spaces as well (any perfect Polish space, for example, or, generalizing from $\omega^*$ to $U(\kappa)$, any space of the form $[0,1]^\kappa$ or $2^\kappa$, if $\kappa$ is regular). | |
| Mar 23, 2016 at 16:33 | history | answered | Lajos Soukup | CC BY-SA 3.0 |