I am trying to understand the structure of the neighborhoods constructed, in Rudin's Proof. This question has been touched upon else where but it really does not go into details of the construction process. Is this how Rudin is constructing his neighborhoods?
$$V_1 = \{\textbf{y} \in P : d(\textbf{x}_1, \textbf{y}) < r \}$$ and subsequently,
$$V_n = \{\textbf{y} \in P: d(\textbf{x}_n, \textbf{y}) < \frac{1}{2} d(\textbf{x}_n, \textbf{x}_{n-1})\}$$
Thanks for the clarification.
To recap his construction: $P$ is perfect in a metric space $(X,d)$. We assume $P = \{x_1, x_2, x_3,\ldots \}$. $V_1$ was any neighbourhood of $x_1$, i.e. take some $r_1 > 0$ such that $V_1 = B(x_1, r_1) = \{x \in X: d(x_1, x) < r_1 \}$. We know of course that $V_1 \cap P \neq \emptyset$.
Now we want to construct $V_2$ (an open set) such that
As $x_1$ is a limit point of $P$, $V_1$ in fact intersects infinitely many points of $P$. Pick $x_i$ where $i > 1$ such that $x_i \in P \cap V_1$. Then as $x_i$ is in $V_1$, for some $s > 0$ we have that the open ball of radius $s$ sits inside $V_1$. We also pick $t = d(x_1, x_i)$. Then set $r_2 = \frac{\min(s,t)}{2}$ and $V_2 = B(x_i, r_2)$. This is open, the factor $\frac{1}{2}$ wrt $s$ ensures that $\overline{V_2} \subseteq V_1$ and being $< t$ ensures that $x_1 \notin V_2$. Also $x_i$ witnesses $V_2 \cap P \neq \emptyset$.
The step from $n$ to $n+1$ is entirely analogous. We can start from any point $x_j$ in $V_n \cap P$, and pick a larger indexed point in the intersection, shrink the radius so that the first and second condition are fulfilled, which can always be done. There is no need for explicit formulae, they only complicate things in this case, I think. The $V_i$ are bals in the whole space, not in $P$ alone, as you suggest. The above is explicit enough.
(As a lemma you can prove that $\overline{B(x,r)} \subseteq \{y: d(x,y) \le r\}$, so that $\overline{B(x,s)} \subseteq B(x,r)$ for any $0 < s < r$ and $x$, which I use above).
