Open ball, closed ball and sphere in the operator norm

Idea. In any normed space the open unit ball is the interior of the closed unit ball, and the closed ball is the closure of the open one. The proof is the "radial scaling" trick.

Step 1: openness and closedness. The norm is $1$-Lipschitz: $|\,\|A\|-\|B\|\,|\le\|A-B\|$, hence continuous. So $U=\{\|A\|<1\}=\|\cdot\|^{-1}((-\infty,1))$ is open and $F=\{\|A\|\le1\}=\|\cdot\|^{-1}((-\infty,1])$ is closed.

Step 2: $\overline U=F$.
- Since $U\subset F$ and $F$ is closed, $\overline U\subset F$.
- Conversely let $\|A\|\le1$. For $t\in(0,1)$, $\|tA\|=t\|A\|<1$, so $tA\in U$, and
$$\|tA-A\|=(1-t)\|A\|\le1-t\longrightarrow0\quad(t\to1^-).$$
Hence $A\in\overline U$.

Step 3: the boundary. For an open set $U$ the boundary is $\partial U=\overline U\setminus U$ (because $U$ equals its interior). Thus
$$\partial U=F\setminus U=\{\|A\|=1\}.$$

Remark.
- Nothing here uses that the norm is the operator norm: the argument works for *any* norm on *any* real normed space (with $U\ne\emptyset$).
- In the metric space $M_n(\mathbb{R})$ one should not confuse "closed ball is closure of the open ball" with the general metric-space statement, which is false for discrete metrics. Here it holds because of the linear structure.