I previously asked [In which topological spaces does the existence of a loop not contractable to a point imply there is a non-contractable simple loop also?](https://mathoverflow.net/questions/338070/in-which-topological-spaces-does-the-existence-of-a-loop-not-contractable-to-a-p)


Given the broad scope of this question I propose this special case as an independent question:



> 
> In a path-connected subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point?
> 
> 
> 



In a subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point?
One-dimensional metric spaces and planar sets do have the property that you're interested in. To explain why this works out in such generality requires a combination of planar topology, continuum theory, and shape theory.


**One-Dimensional Case:** This is pretty classical, going back to work of Curtis and Fort in the 1950's. 


*Proof Sketch.* Suppose $X$ is a one-dimensional metric space. If $X$ contains a simple closed curve, then any loop parameterizing that curve cannot be null-homotopic. A rigorous argument for this requires the [shape injectivity of one-dimensional spaces](https://wildtopology.wordpress.com/2019/08/06/shape-injectivity-of-the-hawaiian-earring-part-ii/). On the other hand, if $X$ contains no simple closed curves, then the image of any loop in $X$ is a dendrite, which is contractible. Thus either $X$ contains a non-contractible simple closed curve or is simply connected.


**Planar Case:** All the ingredients to prove what you want (and more) can be found in 



> 
> H. Fischer, A. Zastrow, The fundamental groups of subsets of closed
>  surfaces inject into their first shape groups, Algebraic and Geometric
>  Topology 5 (2005) 1655--1676.
> 
> 
> 


*Proof Sketch.* Suppose $X\subseteq \mathbb{R}^2$ is path connected and $\alpha:S^1\to X $ is not null-homotopic. Let $Y=\alpha(S^1)\cup\bigcup\{U\mid U\text{ is a bounded path component of }\mathbb{R}^2\backslash \alpha(S^1)\text{ with }U\subseteq X\}$. Fischer and Zastrow show $Y$ is a Peano continuum homotopy equivalent to a “Sierpinski-like” space, i.e. a copy of the Sierpinski Carpet with some, possibly empty, set of (originally deleted) squares filled back in. Since all we want to do is find *some* simple closed curve that doesn’t contract, pick any bounded path component $C$ of $\mathbb{R}^2\backslash Y$ and a point $z\in C\backslash X$. While $C$ is homeomorphic to an open disk, its boundary need not be a simple closed curve. However, you basically have a planar Peano continuum $Y$ surrounding a “hole” at $z$. Since $Y\subseteq X$ and $z\notin X$, any simple closed curve in $Y$ with winding number $1$ around $z$ will do the trick. If you want to create one explicitly, use some planar geometry to cover $Y$ with enough arc-wise connected open (in $Y$) sets to create a finite loop of neighborhoods in $Y$ with winding number $1$ around $z$ and build your curve piece-wise. This simple closed curve can't contract in $X$ since then it would contract in $\mathbb{R}^2\backslash \{z\}$, which clearly is impossible.


