Timeline for Proving that family of sets has non-empty intersection
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 10, 2019 at 1:46 | vote | accept | Doktor Diagoras | ||
| Apr 5, 2019 at 21:15 | comment | added | Doktor Diagoras | Ah... That makes sense, unfortunately. Thank you for your help - dealing with this crazy topological stuff is pure pain for mortal outsiders. Back to thinking then. | |
| Apr 5, 2019 at 20:59 | comment | added | Nate Eldredge | If you're able to prove that your sets $S_i$ are closed in the product topology, great. My guess is that you will find this difficult or impossible; the product topology is just not well suited to such problems. For instance, it is possible for a set to consist only of measurable functions, yet its closure contains functions which are not measurable. So one of the immediate benefits of $L^2$ is that it only has measurable functions to begin with. | |
| Apr 5, 2019 at 20:57 | comment | added | Nate Eldredge | "Closed under a.e. equality" does not imply that you have closed sets in the product topology. Consider for example $\Omega = \mathbb{N}$ with $\mathbb{P}(\{n\}) = 2^{-n}$. The only null set is empty so your "closed under a.e. equality" is trivial. This does not imply that every set is closed in the product topology. | |
| Apr 5, 2019 at 20:51 | comment | added | Doktor Diagoras | Hm... My line of thought was pretty simplistic here: 1. $X$ is compact as every finite set; 2. $S = \prod_{\omega \in \Omega} X$ is compact as product topology; 3. $S_i \subseteq S$ is closed under a.e. equality and so is compact too because of pointwise convergence. What am I missing? | |
| Apr 5, 2019 at 20:25 | comment | added | Nate Eldredge | @DoktorDiagoras: Well, it's good that your sets are closed under a.e. equality, otherwise it wouldn't be well defined to treat your functions as elements of $L^2$. I don't see that it particularly helps you avoid $L^2$. Alaoglu is really not much more than Tychonoff with window dressing. The real issue is coming up with a topology for which Tychonoff would be useful, i.e. which embeds into some product topology. The advantage of the weak topology is that this has already been done for you. | |
| Apr 5, 2019 at 19:56 | comment | added | Doktor Diagoras | I have been digging through Alaoglu's theorem and noticed a thing that can probably save me from going full $L^2$. It happened that all my $S_i \in \mathfrak{S}$ have nice property - if $f' \stackrel{\mathbb{P}}{=} f''$ then $f' \in S_i \Leftrightarrow f'' \in S_i$, or, speaking simply, if function belongs to some set, then such set contains all functions almost equal to it too. In such case, is it possible to prove its compactness through Tychonoff's theorem, or maybe I'm missing something? | |
| Apr 5, 2019 at 17:15 | comment | added | Doktor Diagoras | OK, looks like I need some time to decipher this - I'm specializing in game theory, so stuff like this is usually a bit outside of my scope. | |
| Apr 5, 2019 at 16:57 | history | answered | Nate Eldredge | CC BY-SA 4.0 |