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Introduction :

Let $A$ be the set of non-negative sequences $\{a_n\}$ such that $\sum_{n\geq 1} a_n=1$. Suppose we have a positive sequence $\{a_n\}\in A$. Consider the set $B\subseteq A$ of sequences $\{ b_n \}$ such that $\sum_{n\geq 1} b_n=1$ and $\left\{\frac{b_n}{a_n}\right\}$ is unbounded. Observe that both $B$ and $A\setminus B$ are convex.

For any $\{b_n\}\in B$ there exists a sequence $\{f_n\}$ such that $\sum_{n\geq 1} a_n f_n<\infty$ and $\sum_{n\geq 1} b_n f_n=\infty$. Indeed if we let $n_k$ be such that $k\leq \frac{b_{n_k}}{a_{n_k}}$ and we let $f_{n_k}=\frac{1}{a_{n_k} k^2}$ and $f_n=0$ otherwise, then \begin{align*} \sum_{n\geq 1} a_n f_n &= \sum_{k\geq 1} a_{n_k} f_{n_k}=\sum_{k\geq 1} \frac{1}{k^2}<\infty\\ \sum_{n\geq 1} b_n f_n &= \sum_{k\geq 1} b_{n_k} f_{n_k}=\sum_{k\geq 1} \frac{1}{k^2}\cdot\frac{b_{n_k}}{a_{n_k}}\geq \sum_{k\geq 1} \frac{1}{k}=\infty \end{align*}

It is also clear that if $\{ b_n\}\in A\setminus B$ then there is $x>0$ such that $\left\{ \frac{b_n}{a_n}\right\}$ is bounded by $x$, and so $\sum_{n\geq 1} b_n f_n\leq x\sum_{n\geq 1} a_n f_n$ and so if $\sum_{n\geq 1} a_n f_n<\infty$ then $\sum_{n\geq 1} b_n f_n<\infty$.

All this means that we can write \begin{align*} B &= \left\{ \{b_n\} \in A : \exists \{f_n\}, \sum_{n\geq 1} a_n f_n < \infty = \sum_{n\geq 1} b_n f_n \right\}\\ &=\bigcup_{\{ f_n \}\in F} B_{\{f_n\}} \end{align*}

with $F=\left\{\{f_n\} : \sum_{n\geq 1} a_n f_n < \infty \right\}$ and $B_{\{f_n\}}=\left\{ \{b_n\} \in A : \sum_{n\geq 1} b_n f_n=\infty \right\}$.

Problem :

My question is the following :

Is there a countable subset $G\subset F$ such that $B=\bigcup_{\{f_n\}\in G} B_{\{f_n\}}$?


Attempt :

In case of a negative result, it might be possible to construct a "Cantor's diagonal" like argument by assuming that such a countable set doesn't exists. I have being trying to characterize the inclusion $B_{\{ f_n\}} \subseteq B_{\{ g_n \}}$ by giving an appropriate comparison test on $\{ f_n \}$ and $\{ g_n \}$ without success. Any idea would be very much appreciated.

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  • $\begingroup$ How can you have $\{b_n\}\in B\backslash A$ when $B\subseteq A$? $\endgroup$
    – Dunham
    Commented Dec 19, 2023 at 16:58
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    $\begingroup$ By bad, I meant $A\setminus B$, thank you for the catch, I have edited accordingly. $\endgroup$
    – P. Quinton
    Commented Dec 19, 2023 at 16:59
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    $\begingroup$ The family $F$ consists of sequences of nonnegative real numbers, right? $\endgroup$ Commented Dec 20, 2023 at 1:13
  • $\begingroup$ @AlexRavsky Yes they are non-negative, my bad. $\endgroup$
    – P. Quinton
    Commented Dec 20, 2023 at 7:45

1 Answer 1

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I guess that the family $F$ consists of sequences of nonnegative real numbers, so we assume this.

Recall (see, for instance, [Dou, §3]) that the cardinal $\mathfrak b$ is the smallest size of a family $\mathcal F$ of functions from $\omega$ to $\omega$ such that there is no function $g$ from $\omega$ to $\omega$ such that for each $f\in\mathcal F$, we have $g(m)\ge f(m)$ for all but finitely many $m\in\omega$. The cardinal $\mathfrak b$ is called small, because it is placed between $\omega_1$ and $\frak c$ (see, in particular, [Dou, Theorem 3.1]). But there are models of ZFC with $\mathfrak b<\mathfrak c$, see [Vau]. Also for any regular cardinals $\kappa$ and $\lambda$ with $\omega_1\le\kappa\le\lambda$ it is consistent with ZFC that $\mathfrak b=\kappa$ and $\mathfrak c=\lambda$, see Theorem 5.1 in [Dou].

Suppose that $B=\bigcup_{\{f_n\}\in G} B_{\{f_n\}}$ for some subset $G$ of $F$. We claim that $|G|\ge\mathfrak b\ge\omega_1$. Indeed, multiplying each sequence $\{f_n\}$ of $G$ by a suitable constant, we can provide that $\sum_{n\ge 1} a_nf_n\le 1$. Now for each $m\in\omega$ pick $\hat f(m)\in\omega$ such that $\sum_{n\ge \hat f(m)+1} a_nf_n\le \frac 1{2^m}$. If $|G|<\mathfrak b$ then there is an increasing function $g$ from $\mathbb N$ to $\mathbb N$ such that for each $f\in G$, we have $g(m)\ge \hat f(m)$ for all but finitely many $m$. For each $n\in\mathbb N$ put $b_n=a_n\cdot |\{m\in\omega:g(m)+1\le n\}|$. Multiplying the sequence $\{b_n\}$ by a suitable constant, we can provide that $\sum_{n\ge 1} b_n=1$. Then $\{b_n\}\in B\setminus \bigcup_{\{f_n\}\in G} B_{\{f_n\}}$, a contradiction.

References

[Dou] E.K. van Douwen, The Integers and Topology, in K. Kunen, J. E. Vaughan (eds.), Handbook of Set-Theoretic Topology, Elsevier, 1984, 111--167.

[Vau] J. E. Vaughan. Small uncountable cardinals and topology // in: Open Problems in Topology, ed: J. van Mill and G.M.Reed, Amsterdam: North-Holland, 1990, 195–216.

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    $\begingroup$ Thank you very much for the answer, I am trying to parse it. Did you mean $|G|\ge\mathfrak b\ge\omega_1 $ and $|G|<\mathfrak b$ ? $\endgroup$
    – P. Quinton
    Commented Dec 20, 2023 at 9:01
  • $\begingroup$ I am guessing that in the end you proved that $\{b_n\}\in B_{g_n}\setminus \bigcup_{\{f_n\}\in G} B_{\{f_n\}}$ right ? I am not sure on how to show that $\sum_{n} a_n g(n)$ is bounded and $\sum_{n} b_n g(n)$ is not, but I hope I can figure it out. $\endgroup$
    – P. Quinton
    Commented Dec 20, 2023 at 9:41
  • $\begingroup$ @P.Quinton You are right in your first comment. Thank you for your attention. I am sorry for the misprints. I corrected them. $\endgroup$ Commented Dec 20, 2023 at 10:55
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    $\begingroup$ @P.Quinton Concerning the second comment, the respective expression in the answer is correct. The function $g$ is auxiliary for the constuction of the sequence $\{b_n\}$ . It provides $\left\{\frac{b_n}{a_n}\right\}$ is unbounded, but $\sum_{n\ge 1} b_nf_n<\infty$ for each sequence $\{f_n\}\in G$. $\endgroup$ Commented Dec 20, 2023 at 11:02
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    $\begingroup$ Right, I was stupid, thanks a lot ! looks good to me. $\endgroup$
    – P. Quinton
    Commented Dec 20, 2023 at 13:55

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