Can Isotopic Arcs be Isotoped So That They Don't Intersect Until $t = 1$? Suppose $A, B$ are two disjoint arcs in $X = \mathbb{R}^3$.  An 'arc' is a homeomorphic image of $[0,1]$.  Let $F$ be an ambient isotopy on $X$ carrying $A$ to $B$ - that means there is a homeomorphism $f_1$ on $X$ with $f_1(A) = B$ and $F = \lbrace f_t$ $|$ $0\leq t \leq 1 \rbrace$ is a homotopy from $f_0 = \text{id}_X$ to $f_1$ such that $f_t$ is a homeomorphism for every $t$.
In other words, $A$ can be carried onto $B$ via a continuous family of homeomorphisms of $X$.
The question is: Does this imply the existence of an ambient isotopy $G$ such that $G_t(A) \cap B = \varnothing$ for all $t \neq 1$?  I think this is not true; any of the standard examples of wild arcs seem to be counterexamples, but I'm having trouble showing how to prove it explicitly.  It's not my normal sort of stuff.
Anyone wanna have a go at it, or know a reference?  Surely it's been considered before.
 A: Definition. Given a subset $E\subset R^n$ we say that $A$ can be "pushed-off itself instantly" if there exists an isotopy
$$
F: E\times I\to R^n, f_t=F(\cdot, t), 
$$
such that $f_0=id$  and $f_t(E)\cap E=\emptyset$ for all $t>0$. David Wright in
Wright, David G., Pushing a Cantor set off itself, Houston J. Math. 2, 439-447 (1976). ZBL0332.57002.
proved that a totally disconnected compact subset $E$ in $R^n$ satisfies this property if and only if it is tame, i.e. there is a homeomorphism $R^n\to R^n$ sending $E$ to a subset of a straight line. Since $R^3$ contains wild subsets homeomorphic to the Cantor set (e.g. Antoine discontinuum), the latter cannot be pushed off itself instantly. Now, as I explained here, each totally disconnected compact subset in $R^n$ is contained in an (embedded) topological arc $B$. Taking such $B$  and an arbitrary topological arc $A\subset R^3$ disjoint from $B$ and ambient isotopic to $B$ (say, obtained by a suitable translation of $B$), you obtain the required example.
