# 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?

## 2 Answers

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.

• You said usually one takes the limes superio/inferior, why's that? I know them form other topics abut I can't see why when working with limits of sets one should go with them. Wouldn't you just take both? They might be the same, like in my case or not. Like how does taking the limit here differ from e.g. of a sequence? – xotix Dec 19 '18 at 18:01
• The limit superior and the limit inferior of a sequence of sets may differ from each other. However, when they coincide, it might indeed be natural to say that defines the limit, however this is not common convention. – SmileyCraft Dec 19 '18 at 18:59

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 $$R$$.