Timeline for "Transitivity" of the Stone-Cech compactification
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2010 at 19:35 | comment | added | Greg Graviton | @François: Ah, silly me, thank you. I was so invested in thinking that $\beta(\mathbb{N}\times\mathbb{N})$ must be way more complicated than $\beta\mathbb{N}$ because there is no natural map $\beta\mathbb{N}\times\beta\mathbb{N} \to \beta(\mathbb{N}\times\mathbb{N})$. | |
| Nov 10, 2010 at 18:11 | comment | added | François G. Dorais | @Greg: $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ are both countable discrete spaces, so $\beta\mathbb{N}$ and $\beta(\mathbb{N}\times\mathbb{N})$ are homeomorphic. | |
| Nov 10, 2010 at 15:20 | comment | added | Greg Graviton | I don't understand the negative answer to Q2. Isn't $x$ supposed to be an ultrafilter in $\beta\mathbb{N}$ instead of an ultrafilter in the far larger $\beta(\mathbb{N}\times\mathbb{N})$? | |
| Apr 18, 2010 at 1:01 | comment | added | Terry Tao | Great, thanks! So in particular we see that an ultrapower $*\mathbb{N}$ will neither inject nor surject into $ \beta\mathbb{N}$ unless it is selective, in which case it is at least injective. That seems to clarify the relationship between ultrapowers and the Stone-Cech compactification. | |
| Apr 18, 2010 at 0:58 | vote | accept | Terry Tao | ||
| Apr 16, 2010 at 21:22 | comment | added | François G. Dorais | I just added the selective ultrafilter argument. Note that the existence of selective ultrafilters is independent of ZFC. | |
| Apr 16, 2010 at 21:15 | history | edited | François G. Dorais | CC BY-SA 2.5 |
addendum
|
| Apr 16, 2010 at 20:58 | comment | added | François G. Dorais | I think the answer to Q2 is yes when $x$ is a selective ultrafilter, which is why I had deleted my earlier remark on Q2. | |
| Apr 16, 2010 at 20:57 | comment | added | François G. Dorais | That's exactly right Tom, there are simply too few continuous functions that map $\mathbb{N}\to\mathbb{N}$. | |
| Apr 16, 2010 at 20:55 | history | edited | François G. Dorais | CC BY-SA 2.5 |
addendum
|
| Apr 16, 2010 at 20:50 | comment | added | Tom Church | As I understand François's argument: the cardinality of $\beta\mathbb{N}$ is bigger than the cardinality of the set of continuous functions $\beta\mathbb{N}\to\beta\mathbb{N}$ taking $\mathbb{N}\to\mathbb{N}$. So given $x$ you can find $y$ so that no such function takes $x$ to $y$. | |
| Apr 16, 2010 at 20:38 | comment | added | Jacques Carette | I don't understand your cardinality argument - there are very many ultrafilters, yes, but there are only $2^{\aleph_0}$ choices that each ultrafilter can pick for $y$, as far as I can tell. What am I not seeing? | |
| Apr 16, 2010 at 19:10 | history | edited | François G. Dorais | CC BY-SA 2.5 |
deletion
|
| Apr 16, 2010 at 19:01 | history | answered | François G. Dorais | CC BY-SA 2.5 |