Timeline for Locally compact Hausdorff space that is not normal
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2023 at 7:01 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added a Wayback Machine link for the dead link
|
| Feb 27, 2011 at 17:29 | comment | added | Henno Brandsma | Yes, I agree we'd need to fix the sequences in advance (say we take the convergents for the continued fraction, or the finite truncations of the decimal representations), and then there'd be a hope of finding explicit such closed sets. Else we'd just be doing a counting argument again, like Jones' lemma already is. | |
| Feb 27, 2011 at 14:18 | comment | added | mathahada | I don't think there is a constructive proof like for the Sogenfrey line, at least if the rational sequences are not given beforehand. If we merely consider two disjoint subsets of the irrationals we can arrange for sequences with even denominators to converge to the first set, and for sequences with odd denominators to converge to the other and then basic neighborhoods suffice to separate the sets. | |
| Feb 27, 2011 at 13:15 | comment | added | Henno Brandsma | [continuing] Such spaces are non-normal already by the theorem that normal pseudocompact spaces are countably compact. So if a course would cover these notions, I'd include $\Psi$-space as well. | |
| Feb 27, 2011 at 13:13 | comment | added | Henno Brandsma | So to recap (comments can only be so long...), I don't know of any direct argument (2 closed sets that cannot be separated), but the counting argument is nice enough, and can also be used to show non-normality of $SxS$, where $S$ is the Sorgenfrey line, or for Mrowka's $\Psi$-space (I think the rational sequence topology is called $\Psi$-like, by some authors, as it's very similar, that first space is also pseudocompact and non-countably compact, and needs a MAD family on $\mathbb{N}$, see the <a href="matwbn.icm.edu.pl/ksiazki/fm/fm41/fm41114.pdf">original paper</a>. | |
| Feb 27, 2011 at 13:06 | comment | added | Henno Brandsma | The Jones' lemma argument is a nice counting argument: in a separable space there are at most $\mathfrak{c} = 2^{\aleph_0}$ many real-valued functions (as each function is determined by its restriction to the countable dense subset), and for every subset $B$ of the closed discrete set $A$ we need a distinct function continuous real-valeud $f_B$ that is 1 on $B$ and 0 on $A \setminus B$, both of which are closed and disjoint in the whole space. This does need Urysohn functions and some set theory, but you cannot really do topology without these anyway, so it's useful in a course, I think. | |
| Feb 27, 2011 at 11:57 | comment | added | mathahada | This is an excellent example! Is there an explicit construction of two closed, disjoint sets that cannot be separated in this space? | |
| Feb 27, 2011 at 10:14 | history | edited | Henno Brandsma | CC BY-SA 2.5 |
added 7 characters in body
|
| Feb 27, 2011 at 10:07 | history | answered | Henno Brandsma | CC BY-SA 2.5 |