= Every set of reals is continuum-Suslin
{c}
Assume the <axiom of choice>, and let $A\subseteq\omega^\omega$. Every such $A$ is $2^{\aleph_0}$-Suslin. To see this, use the elements of <Baire space of sequences> as labels. For each $a\in A$, put the node
$$
\bigl((a,a,\ldots,a),\ a\mathbin{\upharpoonright}n\bigr)
$$
of length $n$ into a <tree>, including its prefixes. The label sequence repeats the same entire real $a$ at every position. An infinite branch therefore projects to that particular $a$, rather than merely to a limit of elements of $A$. The projection is exactly $A$.
A bijection between the label set $\omega^\omega$ and a cardinal of size $2^{\aleph_0}$ converts this into the usual <Kappa-Suslin set> representation. The well-ordering of the labels uses choice; without it, the real-labelled tree is still available but an ordinal-sized relabelling needs a separate hypothesis.
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