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
 A: 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.
