Coning makes any connected layer family the link of a ridge-closed family of the same height

Let $F$ be a connected layer family of rank $d\ge1$ on $V$ and add one new symbol $a$. The layers $\{a\cup B:B\in F_t\}\cup\{Q\subseteq V:|Q|=d+1,\ \binom{Q}{d}\subseteq F_t\}$ form a ridge-closed family $G$ of rank $d+1$ on $V\cup\{a\}$ with exactly the same length, and the link of $a$ in $G$ is $F$.

Consequently $h(n,d)\le h_{\rm rc}(n+1,d+1)\le h(n+1,d+1)$: the ridge-closed class is not a restriction of the abstraction but an exact reformulation with the same polynomial exponent. Together with Hirsch.clf_reduces_to_ridge_closed this identifies the polynomial-length question for the EHRR abstraction with the same question for ridge-closed families; and since the EHRR meshes cone without extra bases, ridge-closed families also attain length $\Omega(n^2/\log n)$.

Proof idea. The ridge of $a\cup B$ opposite $a$ is $B$ itself, present in the cone iff $B\in F_t$, so closure adds no base containing $a$; a base $Q\subseteq V$ is added iff all its $d$-subsets are bases of $F_t$, and such $Q$ cannot recur because a fixed $d$-subset cannot be a base twice. Proper shadows are those of the cone, so all subset intervals hold.

Formalization Note The new symbol is none : Option V; the conclusion records the length equality and that every coned base of F lies in the layer of G with the same index.