1550251515-Classical_Complex_Analysis__Gonzalez_

Topology of Plane Sets of Points 103

Intervals of the form [a, a] are clearly connected, and the argument used

above for JR can be applied to any other interval of the real line (open,

closed, semiopen, finite or infinite), provided that the interval be regarded

as a subspace.

To prove that the intervals are the only connected subsets of JR, let C be

any such connected subset, and let a= inf C, b = sup C, where a = -oo

if C is unbounded below, and b = +oo if C is unbounded above. Now,

suppose that a point p such that a < p < b does not belong to C. Then

C C G1 U G2, where

G1 = {x: x E JR,x < p}, G2 = { x : x E JR, x > p}

so that C would be covered by two disjoint open sets. Since the sets

C n G 1 and C n G 2 are not empty, it follows that C is disconnected. This

contradiction shows that C is identical with one of the intervals (a, b ), (a, b],

[a, b) or [a, b], which one depending on whether either of inf C, sup C, or

both, belong or do not belong to C.

Theorem 2.11 A nonempty open set in the plane is connected iff any two

of its points can be joined by a polygonal line all whose points lie in the set.

Proof To show necessity, let A be any open connected set in the complex

plane, and let p be a point in A. Call B 1 the subset of A whose points

can be joined to p by a polygonal lines contained in A, and let B 2 be the

subset of A whose points cannot be joined to p in this manner. The sets

B1 and B2 are both open. In fact, let q E B1 and consider a neighborhood

Ns(q) contained in A (this is possible since A is open). Every point in this

neighborhood can be joined to q by a line segment, and q can be joined top

by a polygonal line. Hence every point of the neighborhood can be joined to
p by a polygonal line, so the whole neighborhood is contained in B1, which
means that B 1 is open. Similarly, for any r in B 2 , consider a neighborhood

Nsi(r) contained in A. If a points in this neighborhood could be joined

to p by a polygonal line lying in A, then r could also be joined top by a

polygonal line lying in A. Since this contradicts the definition of B 2 , no

point in N51 ( r) can be joined to p .by such a polygonal line. It follows that

the whole neighborhood Ns1(r) is contained in B 2 , so B 2 is open. But A is

connected. Hence either B 1 or B 2 is empty. Since p E B 1 , we have B 2 = 0.

Thus all points of A can be joined to p, and given any two points p 1 and
p 2 of A, we can join them by way of p, i.e., by a polygonal line connecting
p 1 to p and then p to p 2 •

To show sufficiency, suppose that A = G 1 U G 2 , where G 1 and G 2 are

disjoint open sets. Let p 1 E G 1 and pz E G 2 , and suppose that P1 and pz

can be joined by a polygonal line contained in A. Then one of the sides of