# Can the set of sequences such that $\frac{b_n}{a_n}$ is unbounded be specified by a countable base of sequences $\{f_n\}$

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.

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

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.

• Thank you very much for the answer, I am trying to parse it. Did you mean $|G|\ge\mathfrak b\ge\omega_1$ and $|G|<\mathfrak b$ ? Commented Dec 20, 2023 at 9:01
• 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. Commented Dec 20, 2023 at 9:41
• @P.Quinton You are right in your first comment. Thank you for your attention. I am sorry for the misprints. I corrected them. Commented Dec 20, 2023 at 10:55
• @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 $\{f_n\}\in G$. Commented Dec 20, 2023 at 11:02
• Right, I was stupid, thanks a lot ! looks good to me. Commented Dec 20, 2023 at 13:55