Interior and boundary of {det > 1}

Idea. The level $\det=1$ is an actual "edge" of the set, because scaling by a factor $t<1$ moves the determinant strictly down.

Let $E=\{\det\ge1\}$ and $O=\{\det>1\}$.

Step 1: $E$ is closed, $O$ is open. Preimages of $[1,\infty)$ and $(1,\infty)$ under the continuous map $\det$.

Step 2: $O\subset\operatorname{int}E$. $O$ is open and $O\subset E$.

Step 3: $\operatorname{int}E\subset O$. Let $A\in E\setminus O$, i.e. $\det A=1$. For $t\in(0,1)$, $\det(tA)=t^n\det A=t^n<1$, so $tA\notin E$; and $\|tA-A\|=(1-t)\|A\|\to0$ as $t\to1^-$. Thus every neighbourhood of $A$ contains a point outside $E$, and $A\notin\operatorname{int}E$. Hence $\operatorname{int}E=O$.

Step 4: the boundary. $E$ is closed, so $\overline E=E$ and
$$\partial E=\overline E\setminus\operatorname{int}E=\{\det\ge1\}\setminus\{\det>1\}=\{\det=1\}=SL_n(\mathbb{R}).$$

Remark.
- The same proof shows $\partial\{\det\ge c\}=\{\det=c\}$ for every $c>0$. For $c=0$ one needs the more delicate SVD argument ($\overline{\{\det>0\}}=\{\det\ge0\}$, problem above), since scaling does not move $\det=0$.
- For the set $\{\det\ge-1\}$ and a point with $\det A=-1$ the scaling must go the other way: $\det(tA)=-t^n>-1$ for $t<1$ stays inside, but for $t>1$ one gets $-t^n<-1$, outside, so such $A$ is a boundary point. Always check the sign.