So I have this statement that I know (I guess I should say 99% sure until I actually see a proof):
Let $X, Y \subset \mathbb{R}^2$ be compact and suppose $F$ is an ambient isotopy carrying $X$ to $Y$. Then there is an ambient isotopy $F'$ carrying $X$ to $Y$ such that $F'$ is constant in a neighborhood of $\infty$.
In other words, if two compact spaces are ambient-isotopic, then you don't need to radically deform the entire plane just to get one to the other. I think it's true for general $\mathbb{R}^n$ with the same argument I'll outline below.
I am having trouble finding a reference for this or proving rigorously that it's true. Can anyone help?
I'm trying to prove it using the Annulus Theorem. My idea for the proof goes as follows. Let $J_1$ be a circle large enough that the trace of $X$ and $Y$ under $F$ are contained in the inside (bounded complementary region, as opposed to outside) of $J_1$, and let $J_2$ be a circle such that the trace of $J_1$ under $F$ is contained in the inside of $J_2$. I want to make it so $F'$ is constant on $J_2$ and its outside, and agrees with $F$ on $J_1$ and its inside.
So basically $F'$ should slide $J_1$ around inside $J_2$. Let $A_t$ denote the annulus between $F_t(J_1)$ and $J_2$. Then $A_0$ is the annulus between $J_1$ and $J_2$ and is homeomorphic to a 'typical' smooth annulus, so has a decomposition into 'transverse arcs'. I want to drag the endpoints of these transverse arcs along $J_1$ according to the action of $F$, and have them stay fixed along $J_2$.
Visually it SEEMS clear that this process gives a continuous extension (it will be bijective by construction), since you can't get "infinite, large-scale winding" in "a compact amount of time," and that's "the only thing that can go wrong." But that's a ton of handwaving.
Please help!
Here is how to do this in the smooth category. Let $F_t(z)$ denote a smooth isotopy, $t\in [0,1]$, $z\in {\mathbb R}^2$; let $K$ denote $$ \{F_t(z): z\in X, t\in [0,1]\}, $$ a compact in the plane. ($F_0=id$, $F_1(X)=Y$.)
Let $B(0,R)$ denote a ball containing $K$. Define the tangent vector field $Z(z,t)=\frac{d}{dt}F_t(z)$ of the flow. Let $\eta(z)$ be a (smooth) cut-off function, which is identically $1$ on $B(0,R)$ and identically $0$ on a ${\mathbb R}^2\setminus B(0,2R)$. Now, define a new vector field $W(z,t)= \eta(z) Z(z,t)$. Since it is identically zero outside of $B(0,2R)$, this vector field defines a flow $G_t(z), t\in [0,1]$ which is constant outside of $B(0,2R)$. By the uniqueness theorem, $G_t(z)=F_t(z)$ for all $z\in B(0,R)$, in particular, all $z\in X$. Hence, $G_t$ is an isotopy taking $X$ to $Y$, which is the identity outside of $B(0,2R)$.
I am sure one can also find an isotopy in the topological category, but that would take more work and I am not sure about the motivation.