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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 nodeof 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.

Use one label for each member of $A$. More explicitly, setAn infinite branch through this tree has a constant first coordinate $a\in A$ and second coordinate $a$, so $p[T_A]=A$. Thus $A$ is $A$-Suslin. Since $A\subseteq {\omega }^{\omega }$ injects into a set of cardinality $2^{{\aleph }_0}$, part i makes $A$ a $2^{{\aleph }_0}$-Suslin set. This proves that Every set of reals is continuum-Suslin.

In ZFC, if $\kappa$ is an infinite cardinal number, every ${\kappa }^{+}$-Suslin set is a union of ${\kappa }^{+}$ many Kappa-Suslin sets. Let a tree $T$ on ${\kappa }^{+}\times \omega$ represent the set. Each branch uses only countably many ordinal labels, so its labels are bounded below ${\kappa }^{+}$, since the successor cardinal is regular in ZFC.

For each $\alpha <{\kappa }^{+}$ restrict $T$ to labels below $\alpha$. Since $|\alpha |\le \kappa$, its projection is a Kappa-Suslin set; taking their union recovers the original projection. The boundedness step requires $\mathrm{cf}({\kappa }^{+})>\omega$. It must not be used without a suitable choice or cofinality assumption in settings involving the axiom of determinacy.