Proof. By removing duplicate elements from the sequence, assume without loss of generality that for each $m, n \in \natp$ with $m \ne n$, $x_{m} \ne x_{n}$.

Let $N \in \natp$. For each $x \in X$, let

Since $x_{1} \in \bigcap_{y \in X}\mathcal{A}(y)$, $1 \in C_{N}(x)$ and $C_{N}(x) \ne \emptyset$. Now, let

be the minimum $n \in C_{N}(x)$ on which the minimal distance from $x$ to $\bracs{x_m|m \in C_N(x)}$ is achieved. Define

(2): For each $x \in X$, $k_{N}(x) \in [N]$, so $I_{N}(x) \in \bracsn{x_n|1 \le n \le N}$.

(3): Let $x \in X$, then by definition of $k_{N}$ and $C_{N}$,

(1, Borel Measurable): Fix $N \in \natp$, then for each $n \in \natp$,

Given that $\mathcal{A}$ is Borel measurable, $\bracs{n \in C_N}= \bracs{x_n \in \mathcal{A}(x)}$ is a Borel set for each $1 \le n \le N$. As a result, $\bracs{J = C_N}$ is Borel for each $J \subset [N]$. Thus $\bracs{j \in C_N, J = C_N}$ is Borel for each $1 \le j \le n$ and $J \subset [N]$.

On the other hand, for each $1 \le n \le N$, the function $x \mapsto d(x, x_{n})$ is continuous and hence Borel measurable. Similarly, for each $J \subset [N]$, the mapping $\real^{J} \to \real$ with $\alpha \mapsto \min_{j \in J}\alpha_{j}$ is also Borel measurable.

The above facts combined show that $\bracs{k_N \le n}$ is a Borel set, and $k_{N}: X \to [N]$ is a Borel measurable function. Now, let

By assumption that $\seq{x_n}$ are distinct, $\bracs{I_N = x_n}= \bracs{k_N = n}$ is a Borel set for each $1 \le n \le N$. Therefore $I_{N}$ is Borel measurable.

(1, $\mathcal{A}$-Admissible): Let $N \in \natp$ and $x \in X$, then

(1, Approximation): Let $x \in X$ and $\eps > 0$. Since $x \in \ol{\mathcal{A}(x)^o}$ and $\seq{x_n}$ is dense in $X$, there exists $N_{0} \in \natp$ such that $x_{N_0}\in \mathcal{A}(x)$ and $d(x, x_{N_0}) < \eps$. By (3),

$\square$