Jordan curve theorem and Alexander horned sphere

Jordan curve or a simple closed curve in the plane \mathbf{R}^2 is the image C of an injective continuous map of a circle into the plane, \varphi:S^1\to\mathbf{R}^2. A Jordan arc in the plane is the image of an injective continuous map of a closed interval into the plane.

With these definitions, the Jordan curve theorem can be stated as follow:

Let C be a Jordan curve in the plane \mathbf{R}^2. Then its complement, \mathbf{R}^2\setminus C, consists of exactly two connected components. One of these components is bounded (the interior) and the other is unbounded (the exterior), and the curve C is the boundary of each component.

Furthermore, the complement of a Jordan arc in the plane is connected.

The Schoenflies theorem is a sharpening of the Jordan curve theorem.

If C\subset\mathbf{R}^2 is a simple closed curve, then there is a homeomorphism f:\mathbf{R}^2\to\mathbf{R}^2 such that f(C) is the unit circle in the plane.

If the curve is smooth then the homeomorphsim can be chosen to be a diffeomorphism. Such a theorem is valid only in two dimension. In three dimensions there are counterexamples such as Alexander’s horned sphere. Although they separate space into two regions, those regions are so twisted and knotted that they are not homeormorphic to the inside and outside of a normal sphere.

Tangent spaces and derivatives

Partial derivatives with respect to local coordinates

Let U\to\mathbf{R}^n be a smooth chart defined over an open set U in M. Then there are smooth real-valued functions x^1,x^2,\dots,x^n on U such that

\displaystyle \varphi(u)=(x^1(u),x^2(u),\dots,x^n(u))

for all u\in U. The functions x^1,x^2,\dots,x^n determined by the chart (U,\varphi) constitute smooth local coordinate functions defined over the domain U of the chart.

Definition Let x^1,x^2,\dots,x^n be smooth local coordinates defined over an open set U in a smooth manifold M of dimension n, and let V be the corresponding open set in \mathbf{R}^n defined such that


Any subspace of a separable metric space is separable

Proposition. Let X be a separable metric space, A be a subpace of X, then A is also separable.,

Proof. let Q\subset X be a countable and dense subset.  For each q\in Q and n\in\mathbf{N}, choose a_{q,n}\in A\cap B(q,1/n) if such intersection is nonempty.  Then, A^*=\bigcup_{q,n} \{a_{q,n}\}  is a countable subset of  A.

Let a\in A, 0<\varepsilon<1. Then choose  q\in B(a,\varepsilon/4)\cap Q.  Now, choose N such that \varepsilon/4\leq N\leq\varepsilon/2.  Then,  B(q,1/N)\cap A\neq\emptyset, so there exists a^*\in A^*, such that d(a^*,q)<1/N.

Now  d(a^*,a)\leq d(a^*,q)+d(q,a)<\varepsilon/2+\varepsilon/4. So a^*\in B(a,\varepsilon)\cap A.  Hence, the closure of A^* in A is equal to A.