Surface of an open ball

Question

In $\mathbb{R}^3$, Euclidean space. Suppose each point is either blue or red. Let $R>0$ be the largest number such that an open ball $B(x_0,R)$ contains only blue points. Is it true that there is at least one limit point of the red points on the surface of this ball?

This is trivial if we are considering real line, but here I'm not so sure. The converse is that for all points on surface we can find an open blue ball centered around it. But there is no way to find an uniform lower bound of the radius of all such balls for all the points--so that I can claim if the statement is not true then I can find $R'>R$, i.e. blue ball can be enlarged. This difficulty does not exist for 1D since there are only two boundary points. What can I do?

Answer 1

If the boundary sphere $S$ contains no limit points of red points $E$, then each blue point $p$ on the sphere has a fixed distance from the red set $\epsilon_p$. So each point $p\in S$ has a ball $B(p,\epsilon_p)$ separate from $E$. This gives an open cover of the sphere. The sphere is compact. Reduce to a finite cover and take the smallest ball. You can enlarge your sphere by this much some amount.

Edit: It is not quite as simple as taking the minimum $\epsilon_p$, look at this picture:

In any case, it is clear we can expand our sphere.

Answer 2

Use compactness. Suppose you have a sequence of red points $p_n$ at radii $d(x_0,p_n) < R + 1/n$. By compactness of $\overline{B(x_0,R+1)}$, some subsequence must converge to a point $p$. By continuity, $d(p,x_0) = \lim_{n \to \infty} d(p_n, x_0) = R$.