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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.

John Samples
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  • My conjecture: let $D$ be the set of wild points of $B$. Then there is no such ambient isotopy if $D$ is non-empty, but if we assume that $A \cap B = \bar{D}$ then there's an isotopy with the desired properties off of $\bar{D}$. – John Samples Jan 22 '21 at 01:17

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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.

Moishe Kohan
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  • Nice, thanks! In that case I guess there is nothing interesting to wonder about, here. – John Samples Feb 01 '21 at 17:19
  • @JohnSamples: There are some interesting variations on this property, for instance, "sticky" subsets, see here. – Moishe Kohan Feb 01 '21 at 17:24
  • Actually I am trying to use that Cantor sets in the plane are tame on this problem, kinda coincidental. See the comments at the end where I sketch a proof idea using that fact (but maybe you know a reference or cleaner proof): https://math.stackexchange.com/questions/4006008/does-mathbbr2-contain-uncountably-many-disjoint-copies-of-the-warsaw-circl – John Samples Feb 01 '21 at 17:35
  • @JohnSamples: I did not think about this one... – Moishe Kohan Feb 01 '21 at 17:59