Timeline for Extracting a common convergent indexing from an uncountable family of sequences
Current License: CC BY-SA 4.0
33 events
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| S Dec 13, 2023 at 1:02 | history | suggested | Peter Mortensen | CC BY-SA 4.0 |
There isn't any need to declare a question - just ask it. Titles are not required to contain question marks (or be actual questions in English - though they could be).
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| Dec 12, 2023 at 19:00 | review | Suggested edits | |||
| S Dec 13, 2023 at 1:02 | |||||
| Dec 12, 2023 at 13:35 | answer | added | terceira | timeline score: 3 | |
| Dec 12, 2023 at 1:33 | comment | added | Jochen Glueck | @Isaac: "I am asking about your use of the symbol $J$ in this specific context." Sorry, I don't understand your question. There is nothing specific about the context here, I simply denoted the index set of the net by $J$ (no special reason for the choice of the letter $J$; it was just the first letter that came to my mind). Of course the index set of the net is not equal to $\mathbb{N}$, in general; if it were, the net would be a sequence. | |
| Dec 12, 2023 at 1:24 | comment | added | Isaac | @JochenGlueck of course I know definition of nets in general. I am asking about your use of the symbol $J$ in the specific context of the posted question. It certainly does not mean $\mathcal{A}$. Do you just mean $J = \mathbb{N}$ here? | |
| Dec 12, 2023 at 0:14 | comment | added | Jochen Glueck | @Isaac: I'd ask you to please look up the definition of nets (for instance in the book I mentioned or on Wikipedia) to see what is meant by the index set of a net. | |
| Dec 12, 2023 at 0:04 | comment | added | Isaac | @JochenGlueck Thank you very much. Just one more thing - what exactly do you mean by $J$ in your notation? | |
| Dec 12, 2023 at 0:01 | vote | accept | Isaac | ||
| Dec 11, 2023 at 14:32 | comment | added | Joel David Hamkins | Another cardinal characteristic! The size of the smallest family of functions such that there is no subfamily of the same size with the common convergent subsequence property. | |
| Dec 11, 2023 at 14:28 | comment | added | Farmer S | @JoelDavidHamkins If CH holds, there is a counterexample of cardinality $2^{\aleph_0}=\aleph_1$: just do a variant of your counterexample, but construct $\left<x_\alpha\right>_{\alpha<\omega_1}$ such that $x_\alpha$ doesn't converge along the first $\alpha$-many index sets, under some enumeration of the index sets in ordertype $\omega_1$. | |
| Dec 11, 2023 at 14:22 | comment | added | Joel David Hamkins | @FarmerS That will be true for regular cardinals above the continuum by pigeon-hole. Not sure about the continuum itself, though. | |
| Dec 11, 2023 at 14:17 | comment | added | Farmer S | Here is a variant question which at least initially might escape the counterexample of @JoelDavidHamkins. Can we find an infinite $I\subseteq\mathbb{N}$ and some $\mathcal{A}'\subseteq\mathcal{A}$ of the same cardinality as $\mathcal{A}$, such that all functions in $\mathcal{A}'$ converge along the index set $I$? | |
| Dec 11, 2023 at 13:59 | answer | added | Farmer S | timeline score: 8 | |
| Dec 11, 2023 at 11:36 | comment | added | Jochen Glueck | Since you asked for a reference: Eric Schechter's "Handbook of Analysis and Its Foundations" has a very detailed treatment of nets and their convergence in Chapter 7. | |
| Dec 11, 2023 at 11:33 | comment | added | Jochen Glueck | Alternatively you can argue as follows: simply choose $(n(j))_{j \in J}$ to be a universal subnet of the sequence $(n)_{n \in \mathbb{N}}$. Then, $(x^\alpha_{n(j)})_{j \in J}$ is, for each $\alpha$, a norm bounded universal net in $X$ and is thus weakly convergent since every closed ball in $X$ is weakly compact and since every universal net in a compact topological space is convergent. | |
| Dec 11, 2023 at 11:29 | comment | added | Jochen Glueck | @Isaac: By a co-final net in $\mathbb{N}$ I mean at net $(n(j))_{j \in J}$ in $\mathbb{N}$ with the following property: for every $m \in \mathbb{N}$ there exists $j_0 \in J$ such that $n(j) \ge m$ for all $j \in J$ that satisfy $j \ge j_0$. What I wrote in my previous comment is an immediate consequence of Tychonoff's theorem (about the compactness of product spaces) and of the general result that a topological space is compact if and only if every net has a convergent subnet. (No separability assumption is needed). | |
| Dec 11, 2023 at 10:36 | comment | added | Isaac | @JochenGlueck Yes, the Banch space is separable. What do you mean by co-final net?? Could you please provide any reference? | |
| Dec 11, 2023 at 9:40 | answer | added | alvoi | timeline score: 9 | |
| Dec 11, 2023 at 7:22 | comment | added | Jochen Glueck | (By the way, may I ask why you assumed the Banach space to be separable?) | |
| Dec 11, 2023 at 7:18 | comment | added | Jochen Glueck | You're probably aware of this, but just to be on the safe side: There does exist a co-final net $(n(j))$ in $\mathbb{N}$ such that the net $(x^\alpha_{n(j)})$ converges weakly for each $\alpha$. | |
| Dec 11, 2023 at 7:04 | comment | added | Isaac | @LSpice A word missing there. Sorry. | |
| Dec 11, 2023 at 7:03 | history | edited | Isaac | CC BY-SA 4.0 |
added 13 characters in body
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| Dec 11, 2023 at 3:33 | comment | added | LSpice | What is missing in "bounded in the norm of $X$ by some for all $n \in \mathbb N$"? | |
| Dec 11, 2023 at 1:16 | answer | added | alvoi | timeline score: 11 | |
| Dec 11, 2023 at 0:48 | history | became hot network question | |||
| Dec 10, 2023 at 22:30 | comment | added | Joel David Hamkins | Yes. continuum $=\frak{c}=2^{\aleph_0}=|\mathbb{R}|=\beth_1$. | |
| Dec 10, 2023 at 21:18 | comment | added | Isaac | @JoelDavidHamkins By size of continuum, you mean the cardinality of $\mathbb{R}$? | |
| Dec 10, 2023 at 20:28 | comment | added | Joel David Hamkins | @terceira That counterexample, like my initial counterexample, has size continuum. But to my way of thinking, the question has become: how big is the smallest counterexample? It is at most the splitting number, which can be strictly less than the continuum. | |
| Dec 10, 2023 at 19:49 | comment | added | terceira | Using the set of all subsets of the natural numbers as the indexing set, then for each such $A$, we let $(x_n^A)$ be the characteristic function of $A$, regarded as a $0,1$ valued sequence in the natural way. | |
| Dec 10, 2023 at 18:20 | comment | added | Pietro Majer | “ If $\mathcal{A}$ is countable, this seems possible by inductive argument”: of course: it is the well known diagonal argument. | |
| Dec 10, 2023 at 17:32 | answer | added | Joel David Hamkins | timeline score: 14 | |
| Dec 10, 2023 at 16:56 | history | edited | Isaac | CC BY-SA 4.0 |
added 48 characters in body
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| Dec 10, 2023 at 16:47 | history | asked | Isaac | CC BY-SA 4.0 |