A fine Euclidean approximation lands in a prescribed tubular neighbourhood

Put $C={\mathbb{R}}^m\setminus U$. If $C=\mathrm{\varnothing }$, take $\epsilon \equiv 1$. It is positive and continuous, and every map into ${\mathbb{R}}^m=U$ already has the required image. This also covers $m=0$. If $M$ is empty the unique error function and the conclusion are vacuous.

Suppose $C\mathrm{\ne }\mathrm{\varnothing }$. Set $\epsilon (p)=\min \{1,\frac{1}{2}d(j(F(p)),C)\}\mathrm{.}$ The distance is a finite real number by [L2]. For each $p$, openness of $U$ gives a ball of some radius $r>0$ about $j(F(p))$ contained in $U$; every point of $C$ is therefore at distance at least $r$. Thus $d(j(F(p)),C)\ge r>0$. By [L2] the distance function is continuous; composing with $j\circ F$ and taking its minimum with $1$ preserves continuity. Explicitly, $\mathrm{\mid }\min (1,a)-\min (1,b)\mathrm{\mid }\le \mathrm{\mid }a-b\mathrm{\mid }$ by the three cases $a,b\le 1$, $a,b\ge 1$, and one on each side. Hence $\epsilon$ is positive, finite, continuous and at most $1$, without making any simultaneous choices of the pointwise radii.

If $\tilde{H}(p)\in C$, the defining infimum gives [step 1.2, L2] $d(j(F(p)),C)\le \mathrm{\parallel }\tilde{H}(p)-j(F(p))\mathrm{\parallel }$. This contradicts the assumed bound, since $\epsilon (p)\le d(j(F(p)),C)\mathrm{/}2<d(j(F(p)),C)$. Thus $\tilde{H}(p)\in U$ for every $p$.

Therefore every approximation with the specified error bound lands in [step 1.1, step 2.1] the prescribed $U$. The construction is choice-free and uses neither an approximation existence theorem nor a new tubular neighbourhood.