208 CHAPTER 8. POSITIVE LINEAR FUNCTIONALS

Proof: Suppose U is a bounded component of JC and U \U is a proper subset of J. LetA⊆ J be the simple curve which contains U \U which is mentioned in Lemma 8.7.16. Letw ∈U .

Let r : R2→ A be the map of Lemma 8.7.16 which fixes all the points of A. Let

f(x)≡{

r(x) if x ∈Ux if x ∈UC

Thus this defines continuous f on all of R2. It is continuous at each point of U and ateach point of UC. If x is in U ∩UC, then r(x) = x and so this is indeed continuous onall of R2. Also, f(x) ̸= w for any x because x ∈U so x /∈ A. Now consider w−R f(x)−w

|f(x)−w|where R is chosen very large, larger than diam

(U). In case f(x) = r(x) , it is not possible

that w−R f(x)−w|f(x)−w| = x because if so, you would have x ∈U and −R f(x)−w

|f(x)−w| = x−w andthe right side is smaller in magnitude than the left. In case f(x) = x, you cannot havew−R f(x)−w

|f(x)−w| = x either because if so, then −R x−w|x−w| = x−w and the two vectors point

in different directions. However, x→ w−R f(x)−w|f(x)−w| is continuous and maps B(w,R) to

B(w,R) and so this would contradict the Brouwer fixed point theorem. Hence U \U = Jas claimed.

Next suppose U is unbounded and let V be a bounded component of JC and let w ∈V .This time let

f(x)≡{

r(x) if x ∈UC

x if x ∈U

As before, this defines continuous f on all of R2. Since w∈V, w is not in J and so f(x) ̸= wfor all x. For R large enough, B(w,R)⊇UC because B(w,R)C ⊆U . If w−R f(x)−w

|f(x)−w| = x for

x ∈UC, you would have −R r(x)−w|r(x)−w| = x−w and this is impossible because |x−w| is less

than R while the left side has magnitude R. If x∈U it is also impossible that w−R f(x)−w|f(x)−w| =

x because this would require that −R x−w|x−w| = x−w and the two vectors point in opposite

directions. Thus x→ w−R f(x)−w|f(x)−w| maps B(w,R) to itself and is continuous but has no

fixed point contradicting the Brouwer fixed point theorem. It follows that U \U = J asclaimed.

Let x(t) ≡ (u(t) ,v(t)) for t ∈ [−1,1] and let y(t) ≡ ( f (t) ,g(t)) , t ∈ [−1,1] and sup-pose u([−1,1]) = [a,b] and g([−1,1]) = [c,d] where x,y are both one to one. The follow-ing picture represents these two curves which lie in the rectangle [a,b]× [c,d] as shownin the picture. Then the conclusion of the following lemma says these two simple curvesintersect.

x(t)

y(t)

Lemma 8.7.18 Let t→ (u(t) ,v(t)) and t→ ( f (t) ,g(t)) for t ∈ [−1,1] be parametriza-tions of two curves in which lie in [a,b]× [c,d] such that u(−1) = a,u(1) = b, and g(−1) =c,g(1) = d. Second component of second curve goes from c to d and first component of