I've recently read some papers and books involving simply connected domains in Euclidean space (dimension at least 2), where domain is an open connected set. The usual definition is a (connected) set for which every continuous closed curve is (freely) contractible while some authors only require that every continuous simple closed curve is contractible. The authors who define simple connectedness using simple closed curves do so in order to use Stokes' Theorem or the Jordan curve theorem somewhere in the sequel; however, they never mention (not even with a reference) that their definition is equivalent to the usual one! My question is if there is a proof written down somewhere (with all the details) proving the equivalence (for domains in $\mathbb{R}^n$ with $n \geq 2$)? If not, does someone know of an "easy" proof using a minimal amount of knowledge, say that of a first course in topology?
[Math] Simple connectedness via closed curves or simple closed curves
at.algebraic-topologygn.general-topology
Related Solutions
A proof of the statement using knowledge that has withstood the test of time is to cite the Alexander duality theorem, for which you can find multiple modern sources. The relevant form of Alexander duality states that the if $X$ is a compact subset of $S^n$, then the reduced Čech cohomology of $X$ is isomorphic to the reduced homology of its complement in the complementary dimension less 1, dimension $q \leftrightarrow n-q-1$. So the reduced $H_0$ of $\mathbb R^2 \setminus X $, which measures disconnectedness and is equal to that of $S^2 \setminus X$, is isomorphic to the 1-dimensional Čech cohomology of $X$; all but 0-dimensional Čech cohomology is trivial for any totally disconnected space, since there are fine covers with 0-dimensional nerve.
The Čech cohomology of $X$ is the direct limit of the cohomology of the nerve of coverings; the nerve is the simplicial complex whose simplices assert that the elements of the cover indexed by its vertices have a non-empty intersection. Using extension theorems, there is a continuous map from a topological space to the nerve of an open cover. What this translates to for any $X$ such that $\mathbb{R}^2 \setminus X$ is disconnected, you can use that fact to define a map from the nerve of any cover of $X$ to $S^1$ which is not null-homotopic, and even when you refine the cover and pull the map back to the refinement, it's still not nullhomotopic. That is inconsistent with $X$ being totally disconnected.
Intuitively, the Alexander theorem detects linking between $X$ and its complement. Every $q$-cycle $Z^q \subset S^n \setminus X$, when $q \ne 0, n$ is the boundary of a $q+1$-chain $C^{q+1}$ in $S^n$, so the nontriviality of $Z^q$ is captured by the way $C^{q+1}$ intersects $X$. Alexander duality is a very natural way to formalize this idea. Ordinary homology is good enough for the complement of $X$: since it is open, it is locally contractible, and the brand of homology or cohomology is not an issue.
The same reasoning shows that the complement of any totally disconnected subset of any manifold of dimension $n \ge 2$ is connected.
The Four Vertex Theorem, due to Mukhopadhyaya in 1909, states that a plane closed simple smooth curve with positive curvature has at least four vertices, where a vertex is a local maximum or minimum of the curvature. There's a detailed discussion in a chapter of Fuchs and Tabachnikov, Mathematical Omnibus.
Best Answer
As pointed out by Pierre and Paul in comments, there are several standard ways to deal with this kind of issue. A good answer really depends what you're assuming you start from, and where you're trying to go to. The Jordan curve theorem and Stoke's theorem are both fairly sophisticated and difficult for beginners to grasp, so it's a bit hard to see how only analyzing embedded curves is streamlining anything, except perhaps helping with people's intuitive images---but even so, it may do more harm than good.
Perhaps it's worth pointing out that this statement is false in greater generality, for instance for closed subsets of $\mathbb R^3$. Here's an example in $\mathbb R^3$: consider a sequence of ellipsoids that get increasingly getting long and thin; to be specific, they can have axes of length $2^{-k}$, $ 2^{-k}$ and $2^k$. Stack them in $\mathbb R^3$ with short axes contained in the $z$-axis, so each one touches the next in a single point with long axes parallel to the $x$-axis, and let $X$ be their union together with the $x$-axis.
Any simple closed curve in $X$ is contained in a single ellipsoid, since to go from one to the next it has to cross a single point, so every simple closed curve is contractible.
However, a closed curve in the $yz$-cross-section that goes down one side and back up the other sides is not contractible. The fundamental group is in fact rather large and crazy.
Anyway, here are some lines of reasoning that can overcome whatever hurdle needs to be ovrcome:
PL approximation, as suggested by Pierre: this is easy, the keyword is "simplicial approximation". I'll phrase it for maps of a circle to Euclidean space as in the question, even though essentially the same construction works in far greater generality. Given an open subset $U \subset \mathbb{R}^n$ and given a map $f: S^1 \to U$, then by compactness $S^1$ has a finite cover by neighborhoods that are components of $f^{-1}$ of a ball. If $U_i$ is a minimal cover of this form, there is a point $x_i$ that is in $U_i$ but not in any other of elements of the cover; this gives a circular ordering to the $U_i$. There is a sequence of points $y_i \in U_i \cap U_{i+1}$, indices taken mod the number of elements of the cover. The line segment between $y_i$ and $y_{i+1}$ is contained in $U_i$, since balls are convex. (This generalizes readily to the statement that for any simplicial complex, there is a subdivision where the extension that is affine on each simplex has image contained in $U$. It also generalizes readily to the case that $U$ is an open subset of a PL or differentiable manifold).
Raising the dimension: if you take the graph of a map of $S^1$ into a space $X$, it is an embedding. If you're (needlessly) worried about integrating differential forms on non-embedded curves, pull the forms back to the graph, where the curve is embedded. If you want to map to a subset of Euclidean space with the same homotopy type, just embed the graph of the map (a subset of $S^1 \times U$ into $\mathbb R^2 \times U$. (There's a very general technique to do this, if the domain is a manifold more complicated than $S^1$, even when it's just a topological manifold, using coordinate charts together with a partition of unity to embed the manifold in the product of its coordinate charts).
The actual issue for integration, using Stoke's theorem etc., is regularity --- to make it simple, restrict to rectifiable curves, and don't worry about embededness. Any continuous map into Euclidean space is easily made homotopic to a smooth curve, by convolving with a smooth bump function---the derivatives are computed by convolving with the variation of the bump, as you move from point to point.
Similarly, you can approximate any continuous map by a real-analytic function, if you convolve with a time $\epsilon$-solution of the heat equation (a Gaussian with very small variance, wrapped around the circle). This remains in $U$ if $\epsilon$ is small enough. A real analytic map either has finitely many double points, or is a covering space to its image; in either case you reduce simple connectivity to the case of simple curves.
Sard's theorem and transversality, as mentioned by Paul. Sard's theorem is nice and elegant and has many applications, including the statement that a generic smooth map of a curve into the plane is an immersion with finitely many self-intersection points, as is any generic smooth map of an $n$-manifold into a manifold of dimension $2n$. If the target dimension is greater than $2n$, then a generic smooth map is an embedding.