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S Oct 6, 2023 at 15:05 history bounty ended CommunityBot
S Oct 6, 2023 at 15:05 history notice removed CommunityBot
Oct 1, 2023 at 8:54 history edited CoffeeArabica CC BY-SA 4.0
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Sep 29, 2023 at 10:40 history edited CoffeeArabica CC BY-SA 4.0
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S Sep 28, 2023 at 13:42 history bounty started CoffeeArabica
S Sep 28, 2023 at 13:42 history notice added CoffeeArabica Draw attention
Sep 27, 2023 at 4:02 comment added plm Thank you. So i was right to expect being mistaken. :) So the most straightforward approach to fix that would be to pick some countable set of pairs of $\tau$ and $p$ that would yield the same intersection. Let's assume that we can restrict the weights $\tau_2$ to monic monomials. Perhaps we can also restrict to countably many sequences $(p_i)_i$, each $p_i$ a sequence, with $\left(\frac{1}{p_i}\right)_i$ a basis of square-summable sequences. Im sorry i don't feel like working it out now. I probably should not have commented in the first place if i was not going to do it right. Sorry.
Sep 26, 2023 at 17:27 comment added CoffeeArabica Your family of intersections is uncountable tho. Since the index set is uncountable..
Sep 26, 2023 at 14:29 comment added plm If i'm not mistaken open subsets of $\mathcal{F}_{\text{fin}}(\mathcal{D})$ are intersections of $\mathcal{F}_{\text{fin}}(\mathcal{D})$ with some $\mathcal{F}(H_\tau,p)$; each of the latter spaces is separable as a subspace of a direct sum of separable metric spaces (the $n$th symmetric power). Then we can take the union of countably many dense sequences, one in each $\mathcal{F}(H_\tau,p)\cap\mathcal{F}_{\text{fin}}(\mathcal{D})$, which is dense in $\mathcal{F}_{\text{fin}}(\mathcal{D})$ and still countable. -This is superficial, as i don't know the background and may easily be mistaken.
Sep 26, 2023 at 13:41 history asked CoffeeArabica CC BY-SA 4.0