Is this an equivalent characterization of rapid filters?

Question

A filter $\mathcal F$ on $\omega$ is called rapid filter if for every function $f\in\omega^\omega$ there exists $X\in\mathcal F$ such that $|X\cap f(n)|\le n$ for $n\in\omega$.

In Lemma 4.6.2 in the book T.Bartoszynski H. Judah: Set Theory. On the Structure of Real Line it is claimed that equivalent condition is that for each sequence $\langle \newcommand{\ve}{\varepsilon}\ve_n \colon n\in\omega\rangle$ there exists $X\in\mathcal F$ such that $\sum_{n\in X} \ve_n$ is finite. I will reproduce the proof given in this book below.

My problem is that, if I understand the proof correctly, it is only shows that it is equivalent to this condition: for a non-negative sequence such that $\ve_n\to0$ we have $X\in\mathcal F$ such that $\sum_{n\in X} \ve_n<1$.

It is not immediately clear to me whether the condition $\sum\limits_{n\in X} \ve_n<1$ can be replaced by the weaker condition $\sum\limits_{n\in X} \ve_n<\infty$ and we still get an equivalent notion.

I will copy the relevant excerpt from the book:

Lemma 4.6.2. The following conditions are equivalent:

• (1) $\mathcal F$ is rapid
• (2) there exists $f_0\in\omega^\omega$ such that for every $f \in \omega^\omega$ there exists $X \in \mathcal F$ such that $|X\cap f(n)| < f_0(n)$ for $n\in\omega$, and
• (3) for any sequence $\langle \ve_n \colon n\in\omega\rangle$ such that $\ve_n\to0$, there exists $X\in\mathcal F$ such that¹ $\sum_{n\in X} \ve_n < \infty$.

Proof. $(1)\to(2)$ is obvious.
$(2)\to(3)$ Suppose that $\ve_n\to0$. Find a sequence $\langle \delta_n \colon n\in\omega\rangle$ such that $\sum\limits_{n=1}^\infty f_0(n)\delta_n<+\infty$. Let $f\in\omega^\omega$ be such that $e_k\le\delta_n$ for $k\in[f(n-1),f(n))$. Suppose $X\in\mathcal F$ is such that $|X\cap f(n)|\le f_0(n)$ for $n\in\omega$. Therefore,² $$\sum_{n\in X}\ve_n \le \sum_{n=1}^\infty f_0(n)\cdot \delta_n < \infty$$
$(3)\to(1)$ Suppose that $f\in\omega^\omega$ is an increasing function. Suppose that $X\in\mathcal F$ is such that³ $\sum\limits_{n\in X}\ve_n<1$. It follows that $|X\cap f(n)|\le n$. $\hspace{3cm}\square$

¹ The formulation of the inequality "$\sum_{n\in X} \ve_n < \infty$" and also the proof seem to indicate that the authors only work with $\ve_n\ge0$, but this is not really an issue. We can simply replace $\ve_n$ by $|\ve_n|$.

² We can choose $\delta_n$ in such way that $\sum\limits_{n=1}^\infty f_0(n)\delta_n<1$. In this way we get as equivalent condition the existence of $X\in\mathcal F$ fulfilling $\sum\limits_{n\in X}\ve_n<1$.

³ Here is the place where in the proof uses the assumption $\sum\limits_{n\in X}\ve_n<1$ is used, although (2) only says $\sum\limits_{n\in X}\ve_n<\infty$.

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So my questions are:

• Am I correct in saying that the above proof in fact shows that $\mathcal F$ is rapid if and only if for each non-negative sequence $\ve_n$ there exists $X\in\mathcal F$ such that $\sum\limits_{n\in X} \ve_n<1$?
• Is the same condition with $\sum\limits_{n\in X} \ve_n<\infty$ instead of $\sum\limits_{n\in X} \ve_n<1$ equivalent?

Answer 1

Note that the definition of filter in Bartoszyński & Judah requires that they extend the Fréchet (cofinite) filter.

In this case, removing finitely many elements from a set in the filter results in another set in the filter, and it is easy to see that in there is an $X \in \mathcal{F}$ such that $\sum_{n\in X} \varepsilon_n$ converges, then there is a $Y \subseteq X$ in $\mathcal{F}$ such that $\sum_{n \in Y} \varepsilon_n < 1$.