Density of continuous knots in the plane transversal to some circles

Question

This is an exercise from the book "Knots and Links" by Rolfsen (exercise 6 in section 2C)

Let $\kappa : S^1 \rightarrow \mathbb{R}^2-(0,0)$ be a continuous imbedding. Let $M := \{ x \in \mathbb{R}^2-(0,0) : |x| \in \mathbb{Z} \}$. Let $\epsilon > 0$. Show that there exists an imbedding $\lambda : S^1 \rightarrow \mathbb{R}^2-(0,0) $ satisfying

1. $|\kappa - \lambda | < \epsilon$

2. $\lambda$ and $\kappa$ are homotopic in $\mathbb{R}^2-(0,0)$

3. $\lambda$ anx $\kappa$ have disjoint images

4. $\lambda$ is transversal to $M$

5. The image of $\lambda$ meets $M$ in only finitely many points.

   Here transversality means that each point of intersection has a neighborhood homeomorphic to the plane such that one set (really its intersection with the neighborhood) is mapped to the x axis and the other to the y axis.

Of course in the smooth setting one has tubular neighborhoods and the density of transversal maps to work with. In the purely topological setting I'm completely lost. Thank you for reading my question.

Answer 1

A quick way to get $\lambda$ is to use complex analysis. Let $f$ be a conformal map of the unit disk $D=\{z:|z|<1\}$ onto the domain $\Omega$ bounded by $\kappa(S^1)$. Let $K=\{w\in\Omega: \operatorname{dist}(w,\partial\Omega)\ge \epsilon /2 \}$. The preimage $f^{-1}(K)$ is a compact subset of $D$. Therefore, it is contained in $|z|<r$ for some $r<1$. The embedding $\lambda$ obtained by restricting $f$ to $|z|=r$ satisfies 1 and 3 (for 1, one has to use Caratheodory's theorem which makes $|f(re^{it})-f(e^{it})|$ uniformly small). It automatically satisfies 2 (by straight-line homotopy) if you make $\epsilon$ smaller than the distance from $\kappa$ to the origin $(0,0)$. Now you are in the smooth setting where you can work toward 4 and 5.

An alternative approach is to use a smooth distance-type function, that is, a positive smooth function $\rho$ defined in $\Omega$ such that $\rho(x)\to 0$ as $x\to\partial \Omega$. Such a function can be constructed without Riemann mapping theorem, by using the Whitney decomposition of $\Omega$. See, e.g., Singular integrals by Stein. Almost every level set of $\rho $ is a disjoint union of smooth $1$-manifolds (closed curves). For sufficiently small $\delta$ the set $\rho=\delta$ separates $\partial\Omega$ from the origin; therefore one of its components does the same. This component provides an approximating curve $\lambda$, but unfortunately the task of finding its parametrization that satisfies 1 remains somewhat nontrivial.