Timeline for Why is $\mathbb R^{\mathbb N}$ not high-dimensional enough?

Current License: CC BY-SA 4.0

13 events

when toggle format	what		by	license	comment
Oct 24 at 23:03	vote	accept	Quertiopler
Oct 24 at 15:44	history	edited	Quertiopler	CC BY-SA 4.0	added 144 characters in body
Oct 24 at 15:07	history	edited	Quertiopler	CC BY-SA 4.0	added 144 characters in body
Oct 24 at 14:04	answer	added	Will Sawin		timeline score: 2
Oct 24 at 9:15	history	edited	Quertiopler	CC BY-SA 4.0	added 2381 characters in body
Oct 23 at 23:05	comment	added	Quertiopler		$p$ is a positive integer, possibly depending on $n$.
Oct 23 at 21:09	comment	added	Michael Hardy		What is $p$? $\qquad$
Oct 23 at 19:04	comment	added	Pietro Majer		@PaulSiegel Actually what you wrote is true in the dual space of $\mathbb R^{\mathbb N}$, i.e. the space $\mathbb R^\omega$ of sequences with finite support, as inductive limit $\bigcup_n\mathbb R^n$. In the space $\mathbb R^{\mathbb N}$ the weak topology is the product topology and gives the point-wise convergence….
Oct 23 at 19:03	history	edited	LSpice	CC BY-SA 4.0	Name of paper; crufted link -> DOI
Oct 23 at 16:31	answer	added	Iosif Pinelis		timeline score: 13
Oct 23 at 16:09	comment	added	Paul Siegel		The problem is not sequences / cardinality considerations - it's the topology on the space of random variables under consideration. I think the point here is that a sequence $S_n$ in $\mathbb{R}^\mathbb{N}$ equipped with the weak topology has a limit if and only if all $S_n$ lie in some $\mathbb{R}^N \hookrightarrow \mathbb{R}^\mathbb{N}$. To prove something resembling a CLT you need a topology in which allows convergence of sequences where the dimension gets arbitrarily large.
S Oct 23 at 15:43	review	First questions
Oct 23 at 16:40
S Oct 23 at 15:43	history	asked	Quertiopler	CC BY-SA 4.0