# Existence of an uncountable set of sequence

I'm back with a question really close to this one : Does it always exist an infinite subset of sequences that satisfy this property?

Now I'm asking myself : does it exist an uncountable set that is dominated. The exact question :

Let $$S \subset \mathbb{N}^{\mathbb{N}}$$ be an uncountable set of sequence. Does it exist $$B\subset S$$ such that

• $$B$$ is uncountable

• $$\forall k \in \mathbb{N},\quad \sup\limits_{u\in B} u_k < +\infty$$

(We include the axiom of choice)

And I don't know what kind of tools to use to prove or disprove this property. Thanks

• What is $u_k$ here? – Wojowu Mar 5 '15 at 19:04
• @Wojowu $u$ is a sequence, $u=(u_0,u_1,\dots)$ with each $u_k$ a natural number. – Andrés E. Caicedo Mar 5 '15 at 19:05
• This is a little confused. You're taking the sup of all the $u_k$ for each fixed $k$. So for example if you take $B$ to be the set of all binary sequences, the sup over all $B$ for a given $k$ will be 1. Is that the question you meant to ask? And even if you mean to take the sup of each sequence $u$, if the sequences are binary all the sups will still be 1. And there are uncountably many binary sequences. – user4894 Mar 5 '15 at 19:09
• @user4894 What do you mean? $S$ is given, so you may not get the chance to pick an uncountable subset of $S$ consisting of binary sequences. The question is whether for each uncountable $S$ there is such uncountable subset $B$. (The problem is that the obvious thinning out process does not work: Given $S$, you can find uncountable subsets $S\supset S_0\supset S_1\supset\dots$ where, for each $k$ there is a constant $n_k$ such that $u_k=n_k$ for all $u\in S_k$. But the intersection of the $S_k$ is either a singleton or empty.) – Andrés E. Caicedo Mar 5 '15 at 19:13
• @AndresCaicedo Oh I see. – user4894 Mar 5 '15 at 19:14

The answer is "yes" if and only if the bounding number $\mathfrak{b}$ is larger than $\omega_1$. Here, by $\mathfrak{b}$, I mean the minimum cardinality of a set $S\subseteq \mathbb{N}^\mathbb{N}$ which is unbounded in the $<^*$ order, where $$x <^* y \iff \exists n\in\mathbb{N}\forall m\ge n\;\; x(m) < y(m)$$
$\mathfrak{b}$ can take many values across different models of set theory, so this shows that your question is independent of ZFC.
First, suppose $\mathfrak{b} = \omega_1$. This means there is a set $S$ of sequences in $\mathbb{N}^\mathbb{N}$, of cardinality $\omega_1$, such that for every $s\in \mathbb{N}^\mathbb{N}$, there is some $u\in S$ such that $u <^* s$ does not hold. Since $S$ has cardinality $\omega_1$, and $\mathbb{N}^\mathbb{N}$ is countably-directed under $<^*$, we may construct a sequence $x_\alpha\in\mathbb{N}^\mathbb{N}$ ($\alpha < \omega_1$) such that $x_\alpha <^* x_\beta$ whenever $\alpha < \beta < \omega_1$, and $X = \{x_\alpha\;|\; \alpha < \omega_1\}$ has the same property. Then any uncountable subset of $X$ is also unbounded, and this violates what you want.
On the other hand, suppose $\mathfrak{b} > \omega_1$, and let $S\subseteq\mathbb{N}^\mathbb{N}$ be an uncountable set. Let $T\subseteq S$ be a subset with cardinality $\omega_1$. Since $\mathfrak{b} > \omega_1$, we can find an $x$ such that $t <^* x$ for all $t\in T$. A Pigeonhole argument shows that there exists $k\in\mathbb{N}$ and a finite sequence $\sigma\in\mathbb{N}^k$ such that for all $u$ in some uncountable subset $U$ of $T$, and for all $n\ge k$, $u(n) < x(n)$, and moreover $u$ looks just like $\sigma$ up to $k$. Then $U$ satisfies what you want.