Connectedness in Metric Spaces: Intervals and Continuous Mappings, Study notes of Mathematics

The concept of connectedness in metric spaces through various theorems and examples. It demonstrates that intervals in the real number system are connected sets, and that the continuous image of a connected set is also connected. The document also proves the intermediate value theorem and fixed point theorem as corollaries of the connectedness properties.

Typology: Study notes

Pre 2010

Uploaded on 09/02/2009

koofers-user-xbq-1
koofers-user-xbq-1 🇺🇸

3

(1)

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
ANALYSIS II
Metric Spaces: Connectedness
Defn. A disconnection of a set A in a metric space (X,d) consists of 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) (1/2,1) is disconnected in the real number system.
Theorem. Each interval (open, closed, half-open) I in the real number system is a connected set.
Pf. Let A1,A2 be a disconnection for I. Let aj Aj, j = 1,2. We may assume WLOG that a1 < a2,
otherwise relabel A1 and A2. Consider E1: = {x A1 | x a2 }, then E1 is nonempty and bounded from
above. Let a: = supE1. But a1 a a2 implies a I since I is an interval. First note that by the lemma to
the least upper bound property either a A1 or a is a limit point of A1. In either case, a A1 since A1 is
closed relative to I. Since A1 is also open relative to the interval I, then there is an ε > 0 so that Nε(a)
A1. But then a+ε/2 A1 and is less than a2, which contradicts that a is the sup of E1. ]
Theorem. If A is a connected subset of real numbers (with the standard metric), then A is an
interval.
Pf. Otherwise, there would be a1 < a < a2, with aj A and a does not belong to A. But then O1 : =
(-,a)A and O2 : = (a,)A 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 IR) open sets are just restrictions of these.
Theorem. The continuous image of a connected set is connected.
Pf. If C is a connected set in a metric space X and there is a disconnection of the image f(C), then it can
be `drawn back' to form a disconnection of C : if { O1, O2 } forms a disconnection for f(C), then {
f-1(O1),f-1(O2) } forms a disconnection for C. ]
Corollary. (Intermediate Value Theorem) Suppose f is a real-valued function which is
continuous on an interval I. If a1, a2 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.
1 of 2 04/24/2000 10:31 PM
Connectedness in Metric Space http://www.math.sc.edu/~sharpley/math555/Lectures/MetricSpaceConnectedness.html
pf2

Partial preview of the text

Download Connectedness in Metric Spaces: Intervals and Continuous Mappings and more Study notes Mathematics in PDF only on Docsity!

ANALYSIS II

Metric Spaces: Connectedness

Defn. A disconnection of a set A in a metric space (X,d) consists of two nonempty sets A

, A

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) ∪(1/2,1) is disconnected in the real number system.

Theorem. Each interval (open, closed, half-open) I in the real number system is a connected set.

Pf. Let A

1

,A

2

be a disconnection for I. Let a

j

∈ A

j

, j = 1,2. We may assume WLOG that a

1

< a

2

otherwise relabel A

1

and A

2

. Consider E

1

: = {x ∈ A

1

| x ≤ a

2

}, then E

1

is nonempty and bounded from

above. Let a: = supE

1

. But a

1

≤ a ≤ a

2

implies a ∈ I since I is an interval. First note that by the lemma to

the least upper bound property either a ∈ A

1

or a is a limit point of A

1

. In either case, a ∈ A

1

since A

1

is

closed relative to I. Since A

1

is also open relative to the interval I, then there is an ε > 0 so that N

ε

(a) ∈

A

1

. But then a+ε/2 ∈ A

1

and is less than a

2

, which contradicts that a is the sup of E

1

[¯]

Theorem. If A is a connected subset of real numbers (with the standard metric), then A is an

interval.

Pf. Otherwise, there would be a

1

< a < a

2

, with a

j

∈ A and a does not belong to A. But then O

1

,a) ∩

A and O

2

: = (a, ∞

A 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 ⊆ IR) open sets are just restrictions of these.

Theorem.

The continuous image of a connected set is connected.

Pf. If C is a connected set in a metric space X and there is a disconnection of the image f(C), then it can

be `drawn back' to form a disconnection of C : if { O

1

, O

2

} forms a disconnection for f(C), then {

f

(O

1

),f

(O

2

) } forms a disconnection for C.

[¯]

Corollary. (Intermediate Value Theorem) Suppose f is a real-valued function which is

continuous on an interval I. If a

, a

∈ I and y is a number between f(a

) and f(a

), then there

exists a between a

and a

such that f(a) = y.

1 of 2 04/24/2000 10:31 PM

Connectedness in Metric Space http://www.math.sc.edu/~sharpley/math555/Lectures/MetricSpaceConnectedness.html

Pf. We may assume WLOG that I = [a

1

, a

2

]. We know that f(I) is a closed interval, say I

1

. Any number y

between f(a

1

) and f(a

2

), belongs to I

1

and so there is an a ∈ [a

1

,a

2

] such that f(a) = y.

[¯]

Corollary. The continuous image of a closed and bounded interval [a,b] is an interval [c,d]

where

c = min

a ≤

x ≤

b

f(x)

d = max

a ≤

x ≤

b

f(x).

Corollary. (Fixed Point Theorem) Suppose that f:[a,b] → [a,b] is continuous, then f has a fixed

point, i.e. there is an α ∈ [a,b] such that f(α) = α.

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 α ∈ [a,b] such that g(α) = 0. This implies that f(α) = α.

[¯]

Robert Sharpley Feb 21 1998

2 of 2 04/24/2000 10:31 PM

Connectedness in Metric Space http://www.math.sc.edu/~sharpley/math555/Lectures/MetricSpaceConnectedness.html