Does every nonconstant curve admit a nonconstant simple subcurve?

Question

The Wikipedia page on the inscribed square problem mentions that an analog of the problem involving triangles instead of squares does hold. However, it doesn't mention anything about triangles inscribed in arbitrary, possibly non-simple, closed plane curves, and I couldn't find any results online for arbitrary closed plane curves either. Naturally, the result is trivial for all closed plane curves that admit some Jordan subcurve, so I started wondering whether all closed plane curves admit Jordan subcurves, but I had no luck finding a proof or a counterexample.

So, I originally posed five questions. The first question was whether nonconstant closed plane curves always admit nonconstant Jordan subcurves, and the second was an analog of the question for general $\mathbb{R}^{n}$. MoisheKohan pointed out in the comments that a closed plane curve whose image is an interval is a counterexample to both. As such, I am now posing a separate but related question in addition to the three other unresolved questions below (still ordered in ascending generality.)

Modified question on closed Euclidean curves: what are some nontrivial necessary or sufficient conditions on nonconstant closed Euclidean curves $\gamma : [0,1] \rightarrow \mathbb{R}^{n}$ with $\gamma(0) = \gamma(1)$ to guarantee the existence of a nonconstant simple closed subcurve $\gamma_{s} : [0, 1] \rightarrow \mathbb{R}^{n}$ with $Im(\gamma_{s}) \subseteq Im(\gamma)$ and $\gamma_{s}(0) = \gamma_{s}(1)$? (e.g. if the set of points where $\gamma$ is injective has a subset of points that is dense-in-itself, then is that sufficient?)

For open Euclidean curves on closed intervals: let $\gamma : [0, 1] \rightarrow \mathbb{R}^{n}$ be an arbitrary Euclidean path satisfying $\gamma(0) \neq \gamma(1)$. Does there exist a nonconstant simple subcurve $\gamma_{s} : [0, 1] \rightarrow \mathbb{R}^{n}$ such that $Im(\gamma_{s}) \subseteq Im(\gamma)$ and $\gamma_{s}(0) = \gamma(0), \gamma_{s}(1) = \gamma(1)$?

For general Euclidean curves: let $I \subseteq \mathbb{R}$ be an arbitrary interval, and $\gamma : I \rightarrow \mathbb{R}^{n}$ be an arbitrary Euclidean curve. If $\gamma$ is nonconstant, then does there exist an interval $I_{s} \subseteq \mathbb{R}$ and a nonconstant simple subcurve $\gamma_{s} : I_{s} \rightarrow \mathbb{R}^{n}$ such that $Im(\gamma_{s}) \subseteq Im(\gamma)$?

For general curves: same statement as the general Euclidean curve question, just with an arbitrary topological space $X$ replacing $\mathbb{R}^{n}$.

Answer 1

For the simple closed curves question, let $A\subseteq [0,1]$ denote the points $t\in [0,1]$ on which $\gamma$ is injective, i.e., $f^{-1}(\{f(t)\})=\{t\}$.

One sufficient condition is that if $A$ has any interior points, then certainly we can construct a simple closed curve in $\text{Im}(\gamma)$, since after reparametrizing so that $(0,\frac{1}{2}]\subseteq A$, we can apply the path-connectedness, hence arc-connectedness, of $\text{Im}(\gamma|_{[\frac{1}{2},1]})$ to obtain an arc $\gamma'$ joining $\gamma(\frac{1}{2})$ and $\gamma(1)$ whose image lies inside of $\text{Im}(\gamma|_{[\frac{1}{2},1]})$, and concatenating $\gamma'$ with $\gamma|_{[0,\frac{1}{2}]}$ then gives a simple closed curve inside of $\text{Im}(\gamma)$.

On the question of whether it's enough for $A$ to merely be dense in itself, I strongly believe the answer is no (nor is it even enough to be dense, period), though I haven't written up a counterexample carefully (see the remark for a rough sketch).

The answer for open and general curves in Euclidean spaces, as well as Hausdorff spaces in general, is yes, since path connectedness implies arc connectedness for Hausdorff spaces. If you drop Hausdorff then for a counterexample (even to a relaxed definition of "simple" that simply requires a curve to not be self-intersecting), let $X$ be any set with cardinality less than the reals, equipped with the trivial topology, then there are many nonconstant paths into $X$, but all are self-intersecting.

Remark

In general, to get a counterexample of a closed curve, injective on a dense set, for which the image contains no simple closed curve, it is easy to see we need an image that is uniquely arcwise connected. A likely counter-example would be a fractal tree-like set, given in the plane for example by starting at stage $0$ with $S_0:=[-1,1]\times \{0\}\cup \{0\}\times [-1,1]$, and then at each stage $n$, constructing $S_n$ by adding a scaled copy of $S_0$ by a factor of $\frac{1}{4^n}$ centered at the midpoint of each segment in $S_{n-1}$. The closure of $\bigcup_n S_n$ is then the image of our curve.

I think the parametrization of this set by a closed curve should result in a dense set of points for which $\gamma$ is injective (including, certainly, the pre-images of the extreme points at each stage of the construction). Moreover, I believe if the parametrization is done in a fairly straightforward way, the curve will actually be injective on a set of full measure even.

The parametrization isn't especially complicated (just traverse each self-similar portion of the tree with equal time spent in each branch), but is somewhat cumbersome to write down. The full-measure argument comes from showing the preimage of each finite stage $S_n$ under $\gamma$ has measure $0$, and arguing that only these points can be non-injective.

However, I have not carefully written the details of this (especially the full measure part), and don't know of a reference where this is done, so take it with something of a grain of salt for the time being.