8.7. HARD TOPOLOGY THEOREMS 207

8.7.3 Jordan Curve TheoremThis treatment of the Jordan curve theorem based on the Brouwer fixed point theorem isthe shortest and most direct proof I have seen. It is from [33]. Any errors are mine. HereJ ⊆ R2 will denote a Jordan curve, defined as the homeomorphic image of the unit circlemeaning that J is γ(S1) where γ is one to one and continuous.

To begin with, note that there is exactly one unbounded component of the complementof a Jordan curve or more generally the complement of a compact set. If U,V are bothunbounded components, they both contain the circle ∂B(0,R) for large enough R that J ⊆B(0,R) , and so there is a continuous curve joining any point in U with a point in V sothat the two must be the same component. Also note that for any nonempty set S it has thesame diameter as its closure. Recall that an arc, also called a simple curve, is the one toone continuous image of a closed interval.

Lemma 8.7.14 Let U be an open set. Then if ∂U denotes those points p such that foreach r > 0, B(p,r) contains points of U and points of UC, then ∂U =U \U.

Proof: If p ∈ ∂U, then p /∈U because U is open. If p /∈U then there would be a ballcontaining p which has no points of U and so p would not be in ∂U . Therefore, p ∈U \Uand so ∂U ⊆U \U. If p ∈U \U, then p is a limit point of U and so B(p,r) contains pointsof U for every r > 0. Since p /∈U, every B(p,r) contains points of UC and so p ∈ ∂U . ■

In the following U will be a connected component of JC.

Lemma 8.7.15 Let J = γ∗ where γ : S1→ R2 is one to one and onto and continuous. IfU is a connected component of JC then U \U ⊆ J.

Proof: Suppose x ∈U \U . I want to show that x ∈ J. If x /∈ J, then, since x is not in U,it must be in a different component V ̸=U . But then x cannot be a limit point of U so x ∈ Jas desired. Thus U \U ⊆ J. ■

Lemma 8.7.16 Let γ : [a,b]→ γ∗ = γ ([a,b]) be one to one and continuous. Then thereexists r :R2→ γ∗ such that r(x) = x for all x∈ γ∗. Also if J is a simple closed curve and Kis a proper compact subset of J, then there exists a simple curve A⊆ J such that K ⊆ A⊆ J.

Proof: By the Tietze extention theorem, there is an extention of γ−1 denoted as r̂ whichmapsR2 onto [a,b]. Consider γ ◦ r̂≡ r. Then r(γ (t))≡ γ (r̂(γ (t)))= γ

(γ−1 (γ (t))

)= γ (t)

so r does what it should. It fixes all the points of γ∗.Now consider the second claim. Since K is a compact proper subset of J, you can start

with a point of J which is not in K and following J in either of the two orientations, thereis a first point of K called a and a last point b. Then the simple curve ab denoting motionfrom a to b is a simple curve which contains K.This is A. ■

With the above observation that U \U ⊆ J, it remains to show that in fact U \U = J.First this is shown under an assumption that JC has at least two components, in particular abounded component.

A

wUorVJ

Lemma 8.7.17 Suppose J is a simple closed curve and suppose there exists a boundedcomponent of JC (JC has at least two components). Then for any component U of JC,U \U = J.