Timeline for Is this set $\sigma$-compact in the Wasserstein space?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 9 at 8:34 | vote | accept | J.R. | ||
| Jun 8 at 15:48 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Jun 8 at 14:42 | comment | added | Saúl RM | Let us continue this discussion in chat. | |
| Jun 8 at 14:39 | comment | added | Saúl RM | I think the first condition should be satisfied, you don't get $1+\varepsilon$ but $(1-\varepsilon/n)+\varepsilon/n$ (because there is a coefficient $(1-\varepsilon/n)$ multiplying $\mu$) | |
| Jun 8 at 14:24 | comment | added | J.R. | But the proposed $\mu_n$ are not in $A$ because the first condition isn't satisfied, you only get $\leq K+\epsilon$, and I think that the third one may also be problematic (e.g., when $\mathcal W (\mu,q)=R$, since maybe the $\mu_n$ are outside the ball) | |
| Jun 8 at 14:17 | vote | accept | J.R. | ||
| Jun 8 at 14:27 | |||||
| Jun 8 at 14:17 | comment | added | J.R. | Okay thanks, I'll mull it over | |
| Jun 8 at 14:15 | comment | added | Saúl RM | I am applying the Baire category theorem to the topological space $A$, not to $\mathcal{P}_1(\mathbb{R}^d)$. So the interior of $A$ is the entire $A$. The only thing one needs to check to apply the Baire category theorem to the space $A$ is that it is closed in $\mathcal{P}_1(\mathbb{R}^d)$, so that is is a complete metric space | |
| Jun 8 at 13:56 | comment | added | J.R. | But doesn't this counterexample only work when the set $A$ has a nonempty interior, right? Because $A$ does have an empty interior (use the same $\mu_n$ you used before you modified the example, it's arbitrarily close to $\mu$, but has the wrong mean, so not in $A$, hence $A$ has empty interior). So I think the argument doesn't work :\ | |
| Jun 7 at 13:47 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Jun 7 at 13:44 | comment | added | Saúl RM | You are right. Well, that should be easily fixable, I'll edit | |
| Jun 7 at 13:39 | comment | added | J.R. | I thought I understood, but now I'm not so sure. The proposed $\mu_n$ are no longer in $A$, since they do not have mean $M$ | |
| Jun 7 at 13:08 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Jun 7 at 12:54 | vote | accept | J.R. | ||
| Jun 8 at 13:56 | |||||
| Jun 7 at 12:36 | comment | added | Saúl RM | A form of the Baire category theorem says that any complete metric space is a Baire space, that is, any countable intersection of dense open sets is nonempty, which implies that no countable union of compacts whose complements are dense can be the whole space. To prove that a compact subset $K$ of $A$ has dense complement, it is enough to show that for any point $\mu$ in $A$ there are points in $A\setminus K$ as close as you want to $\mu$. So it would be enough to check that no compact can contain infinitely many of the measures $\mu_n$, and that $\lim_{n\to\infty}W_1(\mu,\mu_n)=\varepsilon$. | |
| Jun 7 at 12:04 | comment | added | J.R. | Thanks for the answer! Could you elaborate a bit on how exactly you use the Baire category theorem? I don't really have a background in topology unfortunately :\ And how does the second paragraph show that compact subsets of $A$ have dense complements? Sorry for being a bit slow... | |
| Jun 7 at 11:45 | history | answered | Saúl RM | CC BY-SA 4.0 |