Surface and an open ball both centered on the origin point

 No — there is nothing “between” the open ball and the surface.

Why not?

Let

• Open ball:
  \[
  B^\circ = \{x : \|x\| < r\}
  \]
  
• Surface (sphere):
  \[
  S = \{x : \|x\| = r\}
  \]
  

A point in \(\mathbb{R}^3\) must satisfy exactly one of the following: \[
\|x\| < r,\quad \|x\| = r,\quad \text{or}\quad \|x\| > r
\]

There is no fourth possibility, so there is no region that lies strictly between \(\|x\| < r\) and \(\|x\| = r\).

Topological interpretation

• The open ball is the interior
  
• The surface is the boundary
  
• Interior and boundary are disjoint
  
• The boundary has zero thickness
  

Formally: \[
\partial B^\circ = S
\] \[
\overline{B^\circ} = B^\circ \cup S
\]

A common source of confusion

You can form a thin shell, for example: \[
r-\varepsilon < \|x\| < r
\]

But this region is still part of the open ball (its points satisfy \(\|x\|not between the open ball and the surface in a set-theoretic sense.

Final answer

✅ No, there is no region between the open ball and the surface.
The surface is the immediate boundary of the open ball, with nothing separating them.

If you’d like, I can explain how this connects to:

• boundaries vs closures
  
• measure zero vs volume
  
• open/closed sets in topology