It seems like most standard definitions of simple curves in $\mathbb{R^2}$ are those without any self-intersections. Moreover, it can be shown that if $C$ is a simple closed curve, then $C$ is homeomorphic to the circle $S^1$.
(All using the subspace topology of the standard Euclidean topology on $\mathbb{R^n}$ for both, that is.)
So suppose a curve $C$ is not simple, so that it has at least one point of self intersection. That is, if $C$ is identified as the image of a continuous function $f: S^1 \rightarrow \mathbb{R}^2$ from the circle to the plane, then there is at least one point $p\in S^1$ at which $f$ fails to be injective.
In this case, can $C$ still be homeomorphic to a circle?
My hunch is that it can not be, and here's what I've come up with so far:
Towards a direct proof, I think that if $q = f(p)$ is such a point of self-intersection on the curve $C$, then the inverse function $f^{-1}$ may fail to be continuous at $q$. Intuitively, it seems like neighborhoods around this point will always contain points from at least two "branches" of $C$, but I'm not sure how to proceed from here.
Using other topological invariants seemed promising – I think deleting the point $q$ breaks $C$ into at least two connected components while deleting its preimage $p$ leaves the circle in one connected component. Does this constitute a valid proof?
Edit: Phrased another way, if we identify knots as continuous images of $S^1$ embedded in $\mathbb{R^3}$, when is a knot homeomorphic to $S^1$?
Best Answer
You might be interested in the Hahn-Mazurkiewicz Theorem. It tells you that there are a lot of weird subsets of $\mathbb{R}^2$ which are "closed curves", or more properly speaking which are the images of a continuous function $f : S^1 \to \mathbb{R}^2$, for example the Sierpinski triangle.
In fact the theorem characterizes exactly which topological spaces (with a few conditions) can be continuous images of $S^1$. And, of course, if the space is not homeomorphic to $S^1$ then the map $f$ will not be injective (as the comment of @LordSharkTheUnknown shows, the converse is not true).
Here's the usual, very general statement of the theorem, using $[0,1]$ as the domain:
Since $[0,1]$ is the continuous image of $S^1$ and $S^1$ is the continuous image of $[0,1]$, we obtain another version of the theorem as an immediate corollary:
You ask only about subsets of the plane, so you can of course specialize this theorem to apply only to such subsets:
That's a lot of weird subsets.
The construction here shows visually that the Sierpinski Curve is a continuous image of $[0,1]$, and hence also of $S^1$.