Timeline for Does $\mathbb{R}$ have a partite subbase?
Current License: CC BY-SA 4.0
8 events
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| May 8, 2020 at 6:47 | comment | added | Dominic van der Zypen | @RamirodelaVega Thanks for your question about the motivation. My questions often arise out of some playfulness. In this case it happened this way. I was reading this article about Ryser's conjecture and wondered whether the concept of "partiteness" that appears on slide 7 can be applied to topological spaces. I quickly realized that nestedness of open sets posed a problem - therefore the question about a subbase | |
| May 8, 2020 at 6:43 | comment | added | Dominic van der Zypen | @JanKyncl That's right - so first I was wondering whether ${\mathbb R}$ had a subbase without nested elements and then found out this to be true, so I could turn towards partiteness | |
| May 8, 2020 at 1:55 | comment | added | Pierre PC | If it can help, it is not too difficult to see that any $P$ in a possible $\mathfrak P$ would be closed discrete, and there would be for all $[-M,M]$ an $\varepsilon>0$ (depending only on $M$) such that the distance between two points of $P\cap [-M,M]$ would be at least $\varepsilon$ (the distance between two consecutive points of $P$ is locally bounded below, uniformly in $P$). | |
| May 8, 2020 at 1:25 | comment | added | Jan Kyncl | A base would contain (an infinite chain of) sets nested by inclusion, but a partite hypergraph has no pair of nested edges. | |
| May 8, 2020 at 1:15 | comment | added | Ramiro de la Vega | May I ask what the motivation for your question is? and why do you ask for a subbase and not for a base? | |
| May 8, 2020 at 0:19 | comment | added | Will Brian | A near miss: $\mathcal S$ is all open intervals of length $1$, and $\mathfrak{P}$ is all shifts of $\mathbb Z$. | |
| May 7, 2020 at 20:36 | comment | added | YCor | The restriction $X\neq\emptyset$ is non-standard, nevertheless it is usual to assume that components of a partition are nonempty. For instance for $X=\{1,2,3\}$ you currently allow both $\{\{1\},\{2,3\}\}$ and $\{\emptyset, \{1\},\{2,3\}\}$ as distinct partitions. Probably you want to remove $X\neq\emptyset$, and add $P\in\mathfrak{P}$ $\Rightarrow$ $P\neq\emptyset$ in the axioms. | |
| May 7, 2020 at 20:26 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |