Timeline for Is $\beta \mathbb{N}$ homeomorphic to its own square?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2017 at 8:53 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
added doi + link
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| Aug 2, 2017 at 12:25 | comment | added | მამუკა ჯიბლაძე | Sorry I had to go for a while. It is this last fact that explains to me everything. Presumably it is well known for very long time. | |
| Aug 2, 2017 at 10:08 | comment | added | M.González | Let us continue this discussion in chat. | |
| Aug 2, 2017 at 9:51 | comment | added | M.González | If $L$ is compact, then the supremum norm $\|f\|_\infty =\sup_{t\in L}|f(t)|$ induces the compact open topology in $C(L)$. | |
| Aug 2, 2017 at 9:45 | comment | added | მამუკა ჯიბლაძე | Yes so that's what I am asking - the way I know it, the topology on $C(L)=C(L,\mathbb R)$ must be the compact open topology, is it the same as the norm topology? | |
| Aug 2, 2017 at 9:43 | comment | added | M.González | The map $V:C(\beta\mathbb{N})\to\ell_\infty$ defined by $V(f)=(f(n))$ is a bijective linear isometry. | |
| Aug 2, 2017 at 9:41 | comment | added | M.González | For $K$ and $L$ compact, the map $U:C(K\times L)\to C(K,C(L))$ defined by $((Uf)(s))(t) =f(s,t)$ is a bijective linear isometry. | |
| Aug 2, 2017 at 9:36 | comment | added | M.González | For $K$ a compact space, $C(K)$ and $C(K,E)$ are endowed with their supremum norms, and the isomorphisms are considered for the topologies associated to this norm. | |
| Aug 2, 2017 at 9:23 | comment | added | მამუკა ჯიბლაძე | Sorry I don't understand why this helps. I mean this: the bijection $C(X\times Y,Z)\approx C(X,C(Y,Z))$ works for certain topology on $C(Y,Z)$. Is it clear that with $Y=\beta\mathbb N$ and $Z=\mathbb R$ one gets the same topology as in that result of Cembranos that you cite? | |
| Aug 2, 2017 at 8:35 | comment | added | M.González | $K$ homeomorphic to $L$ implies $C(K)$ isomorphic to $C(L)$. | |
| Aug 2, 2017 at 8:26 | comment | added | მამუკა ჯიბლაძე | Is it clear that the topology on $C(\beta\mathbb N)$ that makes that isomorphism work is the same as the Banach space topology on $\ell_\infty$? | |
| Aug 2, 2017 at 7:51 | history | answered | M.González | CC BY-SA 3.0 |