 Solution

For a densely defined closed operator $S$, define its lower normThe closed-range theorem givesThereforeand

We next show that $\mu (T(z))$ is accessible through the matrix-entry evaluations. Put $L_n=\mathrm{span}\{e_1,\dots ,e_n\}$ and defineBecause the canonical span is a core for both operators,

For fixed $m,n$ and grid point $z$, formIts entries belong to $\Lambda$. The smallest singular value of $M_{m,n}(z)$ increases as $m\to \infty$ to ${\nu }_n(T(z))$. Applying the same construction to the conjugate-transposed entries gives ${\nu }_n(T(z{)}^{\ast })$. Thus two nested finite-matrix limits determine $\mu (T(z))$.

The assumed gap continuity of $T$ and the identitymake the corresponding lower-norm tests stable under movement of $z$. The rational grids $G_n$ become dense on every bounded disk. Hence an arithmetic algorithm can, on $G_n\cap {\overline{B}}_n(0)$, use the two finite-section levels above and a vanishing rational tolerance to output all grid cells certified byTaking their closures and letting the mesh and tolerance vanish converges in the Attouch--Wets topology toThe strict existential inequality requires two nested limits, with approximants entering from the prescribed side. In the notation of the arithmetic hierarchy this proves

For the spectrum, equality to zero is the countable intersectionUse the preceding two-level pseudospectral procedure with threshold $1/q$, and add an outer limit $q\to \infty$. Outer approximants remove every point with positive lower norm, while gap continuity and density of the grids retain every zero. Truncating to expanding disks and using vanishing mesh again gives Attouch–Wets convergence. The universal outer intersection reverses the one-sided classification and adds one level, proving