Finite Intersections of Open Balls contain Open Balls about each point

Last updated Nov 1, 2022

Let $(M, d)$ be a Metric Space. Let $n \in \mathbb{N}$, $(x_{i}){i=1}^{n} \subset M$, and $(\epsilon{i}){i=1}^{n} \subset [0, \infty)$. Then for each $y \in \bigcap\limits{i \in [n]} B_{\epsilon_{i}}(x_{i})$, there exists $\zeta > 0$ so that $$B_{\zeta}(y) \subset \bigcap\limits_{i \in [n]} B_{\epsilon_{i}}(x_{i})$$

We first show for two Open Balls, then extend by Induction. Let $x, y \in M$ and let $\epsilon, \delta > 0$. Suppose $z \in B_\epsilon(x) \cap B_{\delta}(y)$. Then $d(x, z) < \epsilon$ and $d(y, z) < \delta$. Let $\zeta = \min(\epsilon - d(x, z), \delta - d(y, z))$. Then, if $w \in B_{\zeta}(z)$ then: $$\begin{align*} d(w, x) &\leq d(x, z) + d(z, w)\\ &< d(x, z) + \zeta\\ &\leq d(x, z) + \epsilon - d(x, z)\\ &=\epsilon \end{align*}$$ and $$\begin{align*} d(w, y) &\leq d(y, z) + d(z, w)\\ &< d(y, z) + \zeta\\ &\leq d(y, z) + \delta - d(y, z)\\ &=\delta \end{align*}$$ Thus $w \in B_{\epsilon}(x) \cap B_{\delta}(y)$ and $B_{\zeta}(z) \subset B_{\epsilon}(x) \cap B_{\delta}(y)$.

TODO make this more rigorous Given $n$ Open Balls and a point in their Set Intersection, we can get an Open Ball around it that is contained in the first $n-1$ Open Balls. Then we apply our argument above to find a smaller ball within its intersection with ball $n$. $\blacksquare$