# What happens if we let the wall thickness of a hollow ball go to zero?

Imagen you have a hollow ball $$\Omega$$: (I am unsure how to write it down.)

$$\Omega(r) = B(R) \setminus B(r), \quad R>r$$

whereas $$B(\hat{r}):=\{v\in\mathbb R^3 : |v|\leq \hat{r})\}$$

So basically you have two balls, both centered at the origin and you take the difference, getting a hollow ball.

If we now let the wall-thickness go to zero, i.e.

$$\lim_{r\to R}\Omega(r)$$

What exactly would happen? Do we get a sphere?

First and foremost, the expression $$\lim_{r\to R}\Omega(r)$$ is ill-defined. However, we can talk about the intersection of $$\Omega(r)$$ for all $$r. This does result in the sphere. There are also general notions of limits for sets, but there is no unique best way to define limits for sets. Two very common ones are the limit superior and the limit inferior. In your case, though, both will still result in the sphere.
The most natural way to take a limit of sets where each one contains the ones that come after it is to take the intersection. And the intersection is indeed the sphere with radius $$R$$.