The following is an excerpt from Billingsley's Convergence of Probability Measures. In the example, $R^\infty$ is the countable product space of $R$ given the metric $\rho(x,y)=\sum_i b(x_i,y_i)/2^i$ where $b(x,y)=1 \wedge |x-y|$. $R_f^\infty$ is the class of all sets of the form $\pi_k^{-1}H$ for $H \in R^k$. In this example, it says for $A_\eta = [y:|y_i - x_i | < \eta, i \le k]$, $\partial A_\eta$ consists of the points $y$ such that $|y_i - x_i | \le \eta$ for all $i \le k$, with equality for some $i$. I cannot figure out why this is so, as I thought they should be those $y$ of the form $|y_i - x_i|=\eta$ for $i \le k$. I would greatly appreciate it if anyone could explain this part to me.
It is clear that $y \in \partial A_{\eta}$ implies $|y_i-x_i| \leq \eta$ for all $i \leq k$. If strict inequality holds for every $i$ then we can find $\delta >0$ such that $|z_i-y_i| <\delta$ for $i \leq k$ implies $z \in A_{\eta}$. This means there is an open neighborhood of $y$ which is contained in $A_{\eta}$ contradicting the fact that $y$ is a boundary point. Hence equality must hold for some $i$. Conversely, suppose $|y_i-x_i| \leq \eta$ for al $i \leq k$ and $|y_j-x_j| =\eta$ for some $j$. Then, by replacing $y_j$ by $y_j \pm \epsilon$ with $\epsilon >0 $ sufficiently small (and keeping other $y_i$'s unchanged) we can approximate $y$ points from inside $A$ as well as points from outside $A$, so $y$ is a boundary point.