Connectedness
Handout #6
Defn 1. Adisconnection of a set Ais two nonempty sets A1, A2whose disjoint
union is Aand 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!∪ 1
2,1!is disconnected.
Theorem 1. Each interval (open, closed, half-open) Iis a connected set.
Proof. Let A1, A2be a disconnection for I. Let aj∈Aj, j = 1,2. We may
assume WLOG that a1< a2, otherwise relabel A1and A2. Consider E1:= {x∈
A1|x≤a2}, then E1is nonempty and bounded from above. Let a:= sup E1. But
a1≤a≤a2implies a∈Isince Iis an interval. First note that by the lemma to
the least upper bound property either a∈A1or ais a limit point of A1. In either
case, a∈A1since A1is closed relative to I. Since A1is also open relative to the
interval I, then there is an ǫ > 0 so that Nǫ(a)∈A1. But then a+ǫ/2∈A1and
is less than a2, which contradicts that ais the sup of E1.2
Theorem 2. If Ais a connected set, then Ais an interval.
Proof. Otherwise, there would be a1< a < a2, with aj∈Aand a6∈ A. But then
O1:= (−∞, a)∩Aand O2:= (a, ∞)∩Aform a disconnection of A.2
Theorem 3. The continuous image of a connected set is connected. The continu-
ous image of [a, b] is an interval [c, d] where c= min
a≤x≤bf(x) and d= max
a≤x≤bf(x).
Proof. Any disconnection of the image f([a, b]) can be ‘drawn back’ to form a discon-
nection of [a, b]: if {O1,O2}forms a disconnection for f(I), then nf−1(O1), f−1(O2)o
forms a disconnection for I= [a, b]. 2
Corollary 1. (Intermediate Value Theorem) Suppose fis a real-valued function
which is continuous on an interval I. If a1, a2∈Iand yis a number between f(a1)
and f(a2), then there exists abetween a1and a2such that f(a) = y.
Proof. We may assume WLOG that I= [a1, a2]. We know that f(I) is a closed
interval, say I1. Any number ybetween f(a1) and f(a2), belongs to I1and so there
is an a∈[a1, a2] such that f(a) = y.2
Theorem 4. Suppose that f: [a, b]→[a, b] is continuous, then fhas a fixed point,
i.e. there is an α∈[a, b] such that f(α) = α.
