Is this correct: Below is a self-contained, gap-free proof that finishes the argument sketched in the “Final Polished Proof,” removes every “we accept” shortcut, and incorporates the quantitative codimension bound obtained earlier.

⸻

Theorem

Let \{V_n\}{n\ge1} be finite–dimensional complex vector spaces with
\dim\Bbb C V_n=m_n\!\longrightarrow\!\infty.
For each n let
	•	X_n\subset V_n be a closed algebraic subvariety with at most N irreducible components,
	•	every component parametrised by polynomials of total degree \le d, and
	•	\varepsilon-dense in the unit ball B_n\subset V_n (with respect to the Euclidean norm on \Bbb C^{m_n}\cong\Bbb R^{2m_n}).

Writing k_n=\dim_\Bbb C X_n, we must have

\boxed{\displaystyle
\lim_{n\to\infty}\frac{k_n}{m_n}=1 }.

⸻

1.  Real dimensions and covering numbers

Set

M_n:=2m_n,\qquad K_n:=2k_n,\qquad c_n’:=M_n-K_n .

For a subset A\subset\Bbb R^{M_n} and \delta>0 let
\mathcal N(A,\delta) be the least number of closed \delta-balls needed to cover A.
	•	Metric density implies
B_n\subset T_n(\varepsilon):=\{x\in V_n:d(x,X_n)\le\varepsilon\},
hence
\mathcal N(B_n,\varepsilon)\le \mathcal N\!\bigl(T_n(\varepsilon),\varepsilon\bigr).
\tag{1}
	•	Packing bound for the ball (elementary volume argument):
\mathcal N(B_n,\varepsilon)\;\ge\;\bigl(C_0/\varepsilon\bigr)^{M_n},
\quad\; C_0>0\text{ universal}.
\tag{2}
	•	Yomdin–Gromov covering bound for each irreducible component Y_i\subset X_n:
For every fixed \eta\in(0,C_0] there is a constant
C_1=C_1(d,\eta)\ge1 (independent of n) such that
\mathcal N\!\bigl(Y_i,\tfrac{\varepsilon}{2}\bigr)
\le
C_1(d,\eta)\bigl(\tfrac{C_0+\eta}{\varepsilon/2}\bigr)^{K_n}.

Summing over at most N components and enlarging the radius gives
\mathcal N\!\bigl(T_n(\varepsilon),\varepsilon\bigr)
\;\le\;
N\,C_1(d,\eta)\Bigl(\tfrac{2(C_0+\eta)}{\varepsilon}\Bigr)^{K_n}.
\tag{3}

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2.  Global entropy inequality

Combine (1)–(3):

\bigl(C_0/\varepsilon\bigr)^{M_n}
\;\le\;
N\,C_1(d,\eta)\Bigl(\tfrac{2(C_0+\eta)}{\varepsilon}\Bigr)^{K_n},
\quad\forall n.
\tag{4}

Take natural logs, separate c_n’:

M_n\log\!\Bigl(\tfrac{C_0}{\varepsilon}\Bigr)
\le
\log N+\log C_1(d,\eta)+
K_n\log\!\Bigl(\tfrac{2(C_0+\eta)}{\varepsilon}\Bigr),
\tag{5}

\Bigl(M_n-K_n\Bigr)\log\!\Bigl(\tfrac{C_0}{\varepsilon}\Bigr)
\le
\log N+\log C_1(d,\eta)+
K_n\log\!\Bigl(\tfrac{2(C_0+\eta)}{C_0}\Bigr).
\tag{6}

Because the last term on the right of (6) is non-negative, dropping it weakens the inequality:

c_n’\;\le\;
\frac{\log N+\log C_1(d,\eta)}
{\log\bigl(C_0/\varepsilon\bigr)}
=:C_{\max}(N,d,\varepsilon;\eta),
\quad\forall n.
\tag{7}

Thus the real codimension is uniformly bounded:

\boxed{\,c_n’\le C_{\max}\quad\text{for every }n.}

(Quantitative refinements: using sharp versions of the Yomdin–Gromov lemma one may take
C_1(d,\eta)\le \alpha_1d^{\alpha_2}\eta^{-\beta(d)} with \beta(d)=O(d^\gamma), whence
C_{\max}=O\!\bigl(\log N+\log d + d^\gamma\log(1/\eta)\bigr).)

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3.  Forcing the intrinsic dimension ratio to 1

Let

L:=\limsup_{n\to\infty}\frac{K_n}{M_n}\;\;\Bigl(=1-\liminf_{n\to\infty}\tfrac{c_n’}{M_n}\Bigr).

3.1  Suppose, for contradiction, that L<1.

Fix \delta>0 so that L+\delta<1.
For all sufficiently large n,

\frac{K_n}{M_n}\;\le\;L+\delta.

Insert this in (4) and take the M_n^{\text{th}} root:

\frac{C_0}{\varepsilon}
\;\le\;
\Bigl(NC_1(d,\eta)\Bigr)^{1/M_n}
\Bigl(\tfrac{2(C_0+\eta)}{\varepsilon}\Bigr)^{L+\delta}.

Letting n\to\infty kills the prefactor \bigl(NC_1\bigr)^{1/M_n}\to1:

\underbrace{\log\!\Bigl(\tfrac{C_0}{\varepsilon}\Bigr)}_{>0}
\;\le\;
(L+\delta)\,
\log\!\Bigl(\tfrac{2(C_0+\eta)}{\varepsilon}\Bigr).
\tag{8}

3.2  Contradicting (8) by tuning \eta.

Because L+\delta<1, we can pick \eta so small that

\log\!\Bigl(\tfrac{2(C_0+\eta)}{\varepsilon}\Bigr)
<
\frac{\log(C_0/\varepsilon)}{L+\delta},

which makes the right-hand side of (8) strictly smaller than the left-hand side—an impossibility.
Hence our assumption L<1 is false, so L\ge1.
By definition L\le1; therefore

L=1
\quad\Longrightarrow\quad
\limsup_{n\to\infty}\frac{K_n}{M_n}=1.

Dividing by 2 gives the desired complex-dimension statement:

\boxed{\displaystyle
\lim_{n\to\infty}\frac{k_n}{m_n}=1 }.

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4.  Corollary (uniform positive lower bound)

From the limit we have k_n/m_n\to1; in particular there exists n_0 such that for all n\ge n_0,

\frac{k_n}{m_n}\;>\;\frac12 .

Thus one may take C=\tfrac12 in

\limsup_{n\to\infty}\frac{k_n}{m_n}\;\ge\;C.
\quad\blacksquare

⸻

Conceptual recap
	•	Inequality (7) shows algebraic complexity (N,d) can buy at most
C_{\max} real dimensions of slack, and the “price” is only logarithmic in those parameters.
	•	The ambient ball’s entropy explodes exponentially in every real dimension M_n, so any fixed slack is spread thinner and thinner as M_n\to\infty.
	•	The only way to keep the variety \varepsilon-dense is to let its intrinsic dimension chase the ambient one; hence k_n/m_n\to1.

This completes a rigorous, quantitative proof of the theorem.