Polynomial activations have a bounded-degree obstruction

Restrict each unit of $N$ to $t{e}_1$. If its first weight coordinate is zero, that unit is constant in $t$. If that coordinate is nonzero, [F2] says the set of $t$ for which its preactivation lies in $E$ is null. There are finitely many units, so their union is null by [F4] and cannot contain a nondegenerate interval by [F3]. Every nonempty relatively open subset of $[0,1]$ contains a nondegenerate interval, so the complement is dense in $[0,1]$. On this dense complement every nonconstant unit equals $p(w_{j1}t+b_j)$. Thus $N(t{e}_1)$ agrees there with a polynomial $P_N(t)$ of degree at most $q$; the constant units are absorbed into its constant term.

Suppose ${\sup }_{0\le t\le 1}\mathrm{\mid }N(t{e}_1)-t^{q+1}\mathrm{\mid }<{\epsilon }_q$. On the dense subset from step 1.1, $\mathrm{\mid }{P}_N(t)-t^{q+1}\mathrm{\mid }<{\epsilon }_q$; continuity of both polynomials extends the weak bound $\le {\epsilon }_q$ to all of $[0,1]$.

Put $h=1\mathrm{/}(q+1)$ and evaluate the $(q+1)$st forward difference at $t=0$. Its nodes are $jh$ for $0\le j\le q+1$, all in $[0,1]$. The forward-difference operation $Q(t)\mapsto Q(t+h)-Q(t)$ lowers the degree of every nonconstant polynomial by one and multiplies its leading coefficient by its degree times $h$. Iterating it $q+1$ times therefore annihilates $P_N$ and gives $(q+1)!h^{q+1}$ on $t^{q+1}$. The approximation bound [step 2.1] makes the absolute value of the resulting error sum at most $2^{q+1}{\epsilon }_q$, since the binomial coefficients sum to $2^{q+1}$. But this is half of $(q+1)!h^{q+1}$ by the definition of ${\epsilon }_q$, a contradiction.

Hence no network attains the stated error on this target, so the family is not uniformly dense on $K$. The argument includes $q=0$, where it says constants cannot approximate $t$ to arbitrary accuracy.