About immersion of $ S^1$

Question

I would like to ask if there is immersion $f: S^1 \to R^2$ that can not be $C^0$ approximated by embedding $S^1 \to R^2$. To be precise, If there exists $\varepsilon_0>0$, such that for any smooth embedding $g: S^1 \to R^2$, $$\sup_{x\in S^1}|f(x)-g(x)|\geq \varepsilon_0 \quad ?$$ I guess $f$ may be $\theta\to \exp^{4\pi i\theta}$, or the figure eight immerion, but I do not know how to prove the result. Thank you very much for your help.

Answer 1

Consider an inmersion $\gamma:S^1\to\mathbb R^2$ which is a figure-eight curve, and let $P$ and $Q$ be points in the interior of the two lobes. There is then an $\epsilon>0$ such that any continuous closed curve which is uniformly $\epsilon$-close to $\gamma$ will not go through $P$ and $Q$ and be freely homotopic to $\gamma$ in $\mathbb R^2\setminus\{P,Q\}$.

Now the winding number of $\gamma$ with respect to $P$ and $Q$ are both non-zero and oppossite. To show that no embedding of $S^1$ is close to $\gamma$ it is then enough to show that any empedding of $S^1$ into $\mathbb R^2\setminus\{P,Q\}$ has winding number with respetc to $P$ and $Q$ either both $\geq0$ or both $\leq0$. This follows from Jordan's curve theorem: indeed, the winding number of an embedded $S^1$ with respect to a point $p$ in its complement is either $1$ (if the point is in its interior) or zero (if it is in the exterior), or $-1$ and $0$, depending on the orientation of the curve.