$$\label{eq:feat_6_unif_rand_lab}
    \indfar_{1} = \emptyset,
    \indfar_{2} = \{\randlabfa_1\},
    \indfar_{3} = \{\randlabfa_2,\randlabfa_3,\randlabfa_4\},
    \indfar_{4} = \{\randlabfa_1,\randlabfa_3,\randlabfa_5\},
    \indfar_{5} = \{\randlabfa_1\},
    \indfar_{6} = \{\randlabfa_5\}.$$

\[lem:unif\_rand\_exch\] Give the features of an exchangeable feature allocation $\randfa_{\infty}$ uniform random labels, and let $\indfar_{n}$ be the set of feature labels for features containing $n$. So $\indfar_{n}$ is a random finite subset of $[0,1]$. Then the sequence $(\indfar_{n})_{n=1}^{\infty}$ is exchangeable.

Note that $(\indfar_{n})_{n=1}^{\infty} = g((\randlabfa_{k})_{k}, (U_{k})_{k}, \randfa_{\infty})$ for some measurable function $g$. So, for any finite permutation $\sigma$, we have that $(\indfar_{\sigma(n)})_{n} = g((\randlabfa_{\tau(k)})_{k}, (U_{k})_{k}, \sigma(\randfa_{\infty}))$ where $\tau$ is a finite permutation that is a function of $\rho$, $(U_{k})$, $\sigma$, and $\randfa_{\infty}$. Now $$((\randlabfa_{\tau(k)})_{k}, (U_{k})_{k}, \sigma(\randfa_{\infty}))
        \eqd ((\randlabfa_{k})_{k}, (U_{k})_{k}, \sigma(\randfa_{\infty}))$$ since the iid sequence $(\randlabfa_{k})_{k}$, the iid sequence $(U_{k})_{k}$, and $\randfa_{\infty}$ are independent by construction and $$((\randlabfa_{k})_{k}, (U_{k})_{k}, \sigma(\randfa_{\infty}))
        \eqd ((\randlabfa_{k})_{k}, (U_{k})_{k}, \randfa_{\infty})$$ since the feature allocation is exchangeable and the independence used above still holds. So $$g((\randlabfa_{\tau(k)})_{k}, (U_{k})_{k}, \sigma(\randfa_{\infty}))
        \stackrel{d}{=} g((\randlabfa_{k})_{k}, (U_{k})_{k}, \randfa_{\infty})$$ It follows that the sequence $(\indfar_{n})_{n}$ is exchangeable.

We can recover the full feature allocation $\randfa_{\infty}$ from the sequence $\indfar_{1},\indfar_{2},\ldots$. In particular, if $\{x_{1}, x_{2}, \ldots\}$ are the unique values in $\{\indfar_{1}, \indfar_{2}, \ldots\}$, then the features are $\{\{n: x_{k} \in \indfar_{n}\}: k = 1,2,\ldots\}$. The 