210 CHAPTER 8. POSITIVE LINEAR FUNCTIONALS

Proof: The following picture is to illustrate the proof of the Jordan curve theorem whichfollows.

ba

ĉ

d̂

c

d

l

p

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Llt

Llb Lrb

LrtJt

Jb

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a b

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Now let J be a Jordan curve. Thus J = γ(S1)

where S1 is the unit circle. Since J iscompact, there exists a≡ inf{x : (x,y) ∈ J} and b≡ sup{x : (x,y) ∈ J}> a. Denote thesepoints on J by (a, l′) ≡ a,(b,r′) ≡ b. Thus the first is a point on J farthest to the left andthe second a point farthest to the right. There are also top and bottom points d̂ and ĉrespectively with second components d̂ and ĉ respectively. Let c < ĉ,d > d̂ as shown. ThusJ is contained in [a,b]× [c,d] as shown. Let C be the boundary of this box. Let L denotethe vertical x = a+b

2 from d to c. There are two Jordan arcs whose union is J joined at thepoints a,b. By Proposition 8.7.20, the line L intersects each of these Jordan arcs. Let Jt bethe first of these arcs intersected by this line L at r in moving from top to bottom and let Jbbe the other one. Let q be the smallest point of L∩ J. I claim that q is in Jb. If not, then qis in Jt and is neither a nor b and neither is r since both are on L. Thus the part of Jt whichgoes from r to q does not include the endpoints of Jt ,a,b. Then Proposition 8.7.20 appliedto [a,b]× [q,r] where q,r are the second components of q,r respectively, would imply thatthis part of Jt between r,q must intersect Jb which is impossible because neither a nor bare on this part of Jt . Such an intersection would mean J is not a simple closed curve.

Let p be the top point of L∩Jb which must be below r. Let l be the bottom point of Jt∩Lwhich is above p. This point must exist since otherwise there would be ln ∈ Jt , ln→ p sop ∈ Jt ∩Jb = {a,b} which is impossible on L. Also let q be the last point of J encountered.Thus q is on Jb as mentioned earlier. Let z be the midpoint of l and p. Then z /∈ Jt andz /∈ Jb so z is in some component of JC.

I want to argue that this component which contains z is a bounded component. Whenthis is done, I will show that it is the only bounded component.

If z is in the unbounded component of JC, then there exists a continuous curve η fromz to a point w on C which does not intersect J. Letting l be the straight line between a andb, if w is above l you could modify η by placing w on the top line of C and if w is belowl we could modify η to place w on the bottom line of C. This involves going from z to thefirst point of the bounding box and then out to one of Lrt ,Lrb,Llb,Llt and along one of thoseslanted lines to a point on the top or bottom of C. (The reason for these slanted lines is that