Defn. A disconnection of a set A is two nonempty sets A1, A2 whose disjoint union is A and each is open relative to A. A set is said to be connected if it does not have any disconnections.
Example. The set (0,[1/2]) È([1/2],1) is disconnected.
Theorem. Each interval (open, closed, half-open) I is a connected set.
Pf. Let A1,A2 be a disconnection for I. Let aj Î Aj, j = 1,2. We may assume WLOG that a1 < a2, otherwise relabel A1 and A2. Consider E1: = {x Î A1| x £ a2 }, then E1 is nonempty and bounded from above. Let a: = supE1. But a1 £ a £ a2 implies a Î I since I is an interval. First note that by the lemma to the least upper bound property either a Î A1 or a is a limit point of A1. In either case, a Î A1 since A1 is closed relative to I. Since A1 is also open relative to the interval I, then there is an e > 0 so that Ne(a) Î A1. But then a+e/2 Î A1 and is less than a2, which contradicts that a is the sup of E1. [¯]
Theorem.
If A is a connected set, then A is an interval.
Pf. Otherwise, there would be a1< a < a2, with aj Î A and a\not Î A. But then O1 : = (-¥,a)ÇA and O2 : = (a,¥)ÇA form a disconnection of A. [¯]
Note.
Each open subset of IR is the countable disjoint union of open intervals.
This is seen by looking at open components
(maximal connected sets) and recalling that each open interval contains
a rational. Relatively (with respect to A Í
IR) open sets are just restrictions of these.
Theorem. The continuous image of a connected set is connected. The continuous image of [a,b] is an interval [c,d] where c = mina £ x £ bf(x) and d = maxa £ x £ b f(x).
Pf. Any disconnection of the image f([a,b]) can be `drawn back' to form a disconnection of [a,b]: if { O1, O2 } forms a disconnection for f(I), then { f-1(O1),f-1( O2) } forms a disconnection for I = [a,b]. [¯]
Corollary.
(Intermediate Value Theorem) Suppose f is
a real-valued function which is continuous on an interval I. If a1,
a2 Î I and y is a number between
f(a1) and f(a2), then there exists a between a1
and a2 such that f(a) = y.
Pf. We may assume WLOG that I = [a1, a2]. We know that f(I) is a closed interval, say I1. Any number y between f(a1) and f(a2), belongs to I1 and so there is an a Î [a1,a2] such that f(a) = y. [¯]
Theorem.
Suppose that f:[a,b] ® [a,b] is continuous,
then f has a fixed point, i.e. there is an aÎ
[a,b] such that f(a) = a.
Pf. Consider the function g(x) : = x- f(x) , then g(a) £ 0 £ g(b). g is continuous on [a,b], so by the Intermediate Value Theorem, there is an a Î [a,b] such that g(a) = 0. This implies that f(a) = a. [¯]
Defn. A function f is called Lipschitz if there is an M > 0 such that
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Theorem. Each Lipschitz function is uniformly continuous.
Theorem. Suppose that K is compact and f:K® K is a contraction, then f has a fixed point in K.
Pf. Let x0 be an arbitrary point in K. Define inductively,
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We claim that the sequence { xn}n = 1¥ is convergent to some a Î K. First note that for each n Î IN
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Hence, by induction, for each n Î IN
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We then see that if m > n, then m = n+k where k Î IN and
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and so {xn}n = 1¥ is Cauchy. It must converge to some limit a which will belong to K since K is closed. But f is continuous, so xn+1 = f(xn)® f(a). Notice also that xn+1®a, so a is our fixed point. [¯]
Theorem.
Suppose that f: [a,b]® K is one-to-one,
onto and continuous, then f-1 is continuous.
Pf. (#1) Suppose that g: = f-1 and yn® y0 Î K. There exists unique xn Î [a,b] such that f(xn) = yn, or equivalently, xn = g(yn). If xn\not® x0, then there exists e0 > 0 and a subsequence xnk such that |xnk-x0|³ e0. This sequence in turn has a subsequence which converges in K to some z Î K. We may as well assume that the subsequence is the sequence { xnk}. f is continuous so ynk = f(xnk)® f(z). But then f(z) = y0 = f(x0). f is one-to-one, so z = x0, which is a contradiction, since |xnk-x0|³ e0. [¯]
Pf. (#2) Let O Í [a,b] be relatively open, then (f-1)-1( O) = f( O). Let C be the complement in [a,b] of O, then C is closed and hence compact. Therefore f(C) is compact in K and conseqently it is closed. Its complement in K must then be open. That complement however is f( O). [¯]