Is $\{0,1\}^{\omega_1}$ sequentially compact? It is claimed that an uncountable product of $[0,1]$ is not sequentially compact, e.g. in Wikipedia (I think replacing $[0,1]$ by $\{0,1\}$ doesn't make much difference).
However, the constructions I saw always take an uncountable set of the same cardinality as the reals.
So I wonder whether $\{0,1\}^{\omega_1}$ is also non-sequentially compact, and if yes, whether one can construct explicitly a sequence without convergent subsequence.
Constructing such a sequence in $\{0,1\}^\mathfrak{c}$ requires no choice; does one need some form of choice to say something about $\{0,1\}^{\omega_1}$?
 A: A family $\mathscr{S}\subseteq[\omega]^\omega$ is a splitting family if for each infinite $A\subseteq\omega$ there is an $S\in\mathscr{S}$ such that $A\cap S$ and $A\setminus S$ are both infinite. The splitting number $\mathfrak{s}$ is defined to be the smallest cardinality of a splitting family. It’s easy to prove that $\omega_1\le\mathfrak{s}\le 2^\omega=\mathfrak{c}$. Theorem $6.1$ of Eric K. van Douwen, ‘The Integers and Topology’, in The Handbook of Set-Theoretic Topology, K. Kunen and J.E. Vaughan, eds., North Holland, $1985$, says that 
$$\begin{align*}
\mathfrak{s}&=\min\{\kappa:\{0,1\}^\kappa\text{ is not sequentially compact}\}\\
&=\min\{\kappa:\text{there is a compact space of weight }\kappa\text{ that is not sequentially compact}\}\\
&=\min\{\kappa:\text{there is a ctbly cpt space of weight }\kappa\text{ that is not sequentially compact}\}\;.
\end{align*}$$
If $\kappa$ and $\lambda$ are regular cardinals with $\omega_1\le\kappa\le\lambda$ it is consistent with $\mathsf{ZFC}$ that $\mathfrak{s}=\kappa$ and $2^\omega=\lambda$; this is (part of) Theorem $5.1$ of the same monograph.
In short, it $\{0,1\}^{\mathfrak{c}}$ is never sequentially compact, but $\{0,1\}^{\omega_1}$ can be, consistently.
