# Non-empty intersection of open balls in $R^n$ contain open balls

I want to prove that if the intersection of two open balls about the points $x, y$ (resp.) is non-empty, then there exists a third ball centered at some point $z\in B_{\epsilon 1}(x)\cap B_{\epsilon 2}(y)$ contained in both of them.

Proof. Let $z\in B_{\epsilon 1}(x)\cap B_{\epsilon 2}(y)$. Suppose that $\epsilon 1-d(x,z) \le \epsilon 2-d(y,z)$ (without loss of generality)

Let $\epsilon = \epsilon 1 - d(x,z)$. First we show that the ball, $B$, of radius $\epsilon$ centered at $z$ is contained in $B_{\epsilon 1}$. Let $c\in B_{\epsilon}$, then $d(z, c) \lt \epsilon = \epsilon 1 - d(x,z)$ so $d(x,c)\le d(z,c)+d(x,z) \lt \epsilon 1$. This proves that $c$ is in $B_{\epsilon 1}$.

Next we show that the ball $B$ is contained in $B_{\epsilon 2}$ as well (hence in the intersection of $B_{\epsilon 1}$ and $B_{\epsilon 2}$). $d(z,c) \lt \epsilon 2-d(y,z)$ so $d(y,c)\lt \epsilon 2$ thus $c$ is in $B_{\epsilon 2}$.

Is this ok?

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Since the intersection of any finite number of open sets is open on any topological space, this is also true for metric spaces, clearly. What you are doing is to show a possible radius to an open ball. It looks nice for me. –  Sigur Jul 24 at 22:25
Looks okay, except for a typo in the middle paragraph of the proof, where you should assume that $c\in B_\epsilon$. –  Stefan Hamcke Jul 24 at 22:30
@Stefan. Thanks${}{}{}{}$ –  saadtaame Jul 24 at 22:32
@Sigur This is to show that the balls form a basis for $R^n$. I can't assume that they are open in $R^n$ –  saadtaame Jul 24 at 22:34
@saadtaame, ok, I see. In this case, you are on the right way. Good job. –  Sigur Jul 24 at 22:35
One very small style thought: Instead of WLOGing $\epsilon_1 - d(x,z) \leq \epsilon_2 - d(y,z)$, perhaps set $\epsilon = \min\{ \epsilon_1 - d(x,z), \epsilon_2 - d(y,z)\}$. Then all you have to see is that, for any $c\in B_\epsilon(z)$, \begin{align*} d(c,x) &\leq d(c,z) + d(z,x) \\ &< \epsilon + \epsilon_1 \\ &\leq \epsilon_1\end{align*} and likewise $d(c,y)<\epsilon_2$.