The closure of {det > 0}

Idea. A small perturbation of a singular matrix is invertible, but its determinant could be negative; we must control the *sign*. The SVD makes the sign explicit.

Step 1: $U$ is open and $\overline U\subset\{\det\ge0\}$. $\det$ is continuous, so $U=\det^{-1}((0,\infty))$ is open and $\{\det\ge0\}=\det^{-1}([0,\infty))$ is closed. Since $U\subset\{\det\ge0\}$, taking closures gives $\overline U\subset\{\det\ge0\}$.

Step 2: every matrix with $\det A>0$ is in $U$; remains $\det A=0$. Let $\det A=0$. Write the SVD $A=P\Sigma Q^T$ with $P,Q$ orthogonal and $\Sigma=\operatorname{diag}(\sigma_1,\dots,\sigma_n)$, $\sigma_i\ge0$. Since $A$ is singular, the set $Z=\{i:\sigma_i=0\}$ is non-empty. Let $\delta=\det P\cdot\det Q\in\{\pm1\}$ and let $i_0\in Z$.

For $\varepsilon>0$ define $\Sigma_\varepsilon=\operatorname{diag}(\sigma^{(\varepsilon)}_i)$ by
$$\sigma^{(\varepsilon)}_i=\begin{cases}\sigma_i&i\notin Z,\\ \varepsilon&i\in Z,\ i\ne i_0,\\ \delta\,\varepsilon&i=i_0,\end{cases}$$
and $A_\varepsilon=P\Sigma_\varepsilon Q^T$.

Step 3: the sign. $\det A_\varepsilon=\det P\,\det Q\,\det\Sigma_\varepsilon=\delta\cdot\delta\,\varepsilon^{|Z|}\prod_{i\notin Z}\sigma_i=\varepsilon^{|Z|}\prod_{i\notin Z}\sigma_i>0$ because $\delta^2=1$. So $A_\varepsilon\in U$.

Step 4: closeness. $A_\varepsilon-A=P(\Sigma_\varepsilon-\Sigma)Q^T$ and the operator norm is invariant under orthogonal multiplication, so $\|A_\varepsilon-A\|=\|\Sigma_\varepsilon-\Sigma\|=\varepsilon\to0$. Therefore $A\in\overline U$.

Step 5: the boundary. $U$ is open so $\partial U=\overline U\setminus U=\{\det\ge0\}\setminus\{\det>0\}=\{\det=0\}$.

Numeric check. For $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ one may take $P=I$, $Q=I$ ($\delta=1$) and $A_\varepsilon=\operatorname{diag}(1,\varepsilon)$ with $\det=\varepsilon>0$. For $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, an SVD has $\delta=-1$, and the sign flip is indeed needed.

Remark.
- Symmetrically $\overline{\{\det<0\}}=\{\det\le0\}$, so $\{\det>0\}$ and $\{\det<0\}$ share the same boundary $\{\det=0\}$. Thus $GL_n(\mathbb{R})$ is cut by the singular matrices into two open pieces, the first hint at its two connected components (Part 5).
- The naive perturbation $A+\varepsilon I$ may have negative determinant for both signs of $\varepsilon$: for $A=\operatorname{diag}(1,-1,0,0)$ one gets $\det(A+\varepsilon I)=\varepsilon^2(1+\varepsilon)(\varepsilon-1)<0$ for all small $\varepsilon\ne0$. The SVD argument avoids this issue altogether.