Analysis with an introduction to proof 5th by steven lay ch05a

So, the continuous image of a compact set is compact.The continuous image of a closed set may not be closed.. The Heine-Borel Theorem 3.5.5 says S is compact iff S is closed and bounded.

Chapter 5 Limits and Continuity

Section 5.3 Properties of Continuous Functions

So, the continuous image of a compact set is compact.

The continuous image of a closed set may not be closed.

The Heine-Borel Theorem 3.5.5 says S is compact iff S is closed and bounded.

The continuous image of a bounded set may not be bounded.

But if a set is both closed and bounded, then its image under a continuous function will be both closed and bounded.

Note: A function f : D → is said to be bounded if its range f (D) is

a bounded subset of

So, a continuous function may not be bounded, even if its domain is bounded

Recall from Definition 3.5.1

A set S is compact if every open cover of S contains a finite subcover.

Idea of proof:

f

Open cover of f (D) Open cover of D

Let G = {Gα} be an open cover of f (D).

Since f is continuous, the pre-image of each set in G is open.

These pre-images form an open cover of D.

Since D is compact, there is a finite subcollection that covers D.

The corresponding sets in the range form a finite subcovering of f (D).

Hence f (D) is compact.

Theorem 5.3.2

Let D be a compact subset of and suppose f : D → is continuous

Then f (D) is compact.

Let D be a compact subset of and suppose f : D → is continuous.

Then f assumes minimum and maximum values on D That is, there exist points x1 and x2 in D such that f (x1) ≤ f (x) ≤ f (x2) for all x ∈ D.

Proof:

We know from Theorem 5.3.2 that f (D) is compact.

Lemma 3.5.4 tells us that f (D) has both a minimum, say y1, and a maximum, say y2.

Since y1 and y2 are in f (D), there exist x1 and x2 in D such that f (x1) = y1 and f (x2) = y2

It follows that f (x1) ≤ f (x) ≤ f (x2) for all x ∈ D ♦

If we require the domain of the function to be a compact interval, then the following holds:

Corollary 5.3.3

Let f : [a, b] → be continuous and suppose f (a) < 0 < f (b) Then there

exists a point c in (a, b) such that f (c) = 0.

Proof:

Let S = {x ∈ [a, b] : f (x) ≤ 0} Since a ∈ S, S is nonempty.

Clearly, S is bounded above by b, and we may let c = sup S. We claim f (c) = 0.

Suppose f (c) < 0.

y = f (x)

• •

S U

Then there exists a neighborhood U of c such that f (x) < 0 for all

x ∈ U ∩ [a, b].

And U contains a point p such that c < p < b.

But f ( p) < 0, since p ∈ U, so p ∈ S.

This contradicts c being an upper bound for S.

Lemma 5.3.5

Now suppose f (c) > 0.

y = f (x)

c p

Then there exists a neighborhood U of c such that

f (x) > 0 for all x ∈ U ∩ [a, b].

And U contains a point p such that a < p < c.

Since f(x) > 0 for all x in U, S ∩ [p, c] = ∅

This implies that p is an upper bound for S, and contradicts c being the least upper bound for S.

• •

We conclude that f (c) = 0 ♦

Let f : [a, b] → be continuous and suppose f (a) < 0 < f (b) Then there

exists a point c in (a, b) such that f (c) = 0.

Proof:

Let S = {x ∈ [a, b] : f (x) ≤ 0} Since a ∈ S, S is nonempty.

Clearly, S is bounded above by b, and we may let c = sup S. We claim f (c) = 0.

Lemma 5.3.5

(Intermediate Value Theorem)

Suppose f : [a, b] → is continuous Then f has the intermediate value property

on [a, b] That is, if k is any value between f (a) and f (b) , i.e.

f (a) < k < f (b) or f (b) < k < f (a), then there exists a point c in (a, b) such that f (c) = k.

Proof:

Suppose f (a) < k < f (b) Apply Lemma 5.3.5 to the continuous function

g : [a, b] → given by g(x) = f (x) – k.

Then g(a) = f (a) – k < 0 and g(b) = f (b) – k > 0.

Thus there exists c ∈ (a, b) such that g(c) = 0.

But then f (c) – k = 0 and f (c) = k.

When f (b) < k < f (a), a similar argument applies to the function g(x) = k – f (x) ♦

Theorem 5.3.6

(Intermediate Value Theorem)

Suppose f : [a, b] → is continuous Then f has the intermediate value property

on [a, b] That is, if k is any value between f (a) and f (b) , i.e.

f (a) < k < f (b) or f (b) < k < f (a), then there exists a point c in (a, b) such that f (c) = k.

The idea behind the intermediate value theorem is simple when we graph the function

y = f (x)

f (b)

y = k

f (a)

If k is any height between f (a) and f (b), there must be some point c in (a, b) such

that f (c) = k.

That is, if the graph of f is below y = k

at a and above y = k at b, then for f to be continuous on [a, b], it must cross y = k

somewhere in between

Theorem 5.3.6

(an application in geometry)

Let C be a bounded closed subset of the plane We claim there exists a square S that circumscribes C That is, C ⊆ S and each side of S touches C

C

Given any α ∈ [0, 2π], let r be a ray from the origin having angle α with the positive x axis.

This ray determines a unique circumscribing rectangle whose sides are parallel and

perpendicular to r.

Let A(α ) be the length of the sides parallel to r.

A(α )

Let B(α) be the length of the sides

perpendicular to r.

Now define f : [0, 2π] → by

f (α ) = A(α) – B(α)

If for some α the circumscribing rectangle

is not a square, then A(α) ≠ B(α), and we

suppose f (α) = A(α) – B(α) > 0

r

α

Example 5.3.8

We have f (α) = A(α) – B(α) > 0 Now let β = α + π/2

r

β

Then the circumscribing rectangle

is unchanged except the labeling

of its sides is reversed

B(β )

A(β ) So f (β ) = A(β ) – B(β ) < 0

Assuming that f is a continuous function,

the Intermediate Value Theorem implies there exists some angle θ with α < θ < β

such that f (θ) = 0

But then A(θ) = B(θ) and the circumscribing rectangle for this angle is a square

Let C be a bounded closed subset of the plane We claim there exists a square S that circumscribes C That is, C ⊆ S and each side of S touches C

Given any α ∈ [0, 2π], let r be a ray from the origin having angle α with the positive x axis.

Example 5.3.8

β

α

Let C be a bounded closed subset of the plane We claim there exists a square S that circumscribes C That is, C ⊆ S and each side of S touches C

Example 5.3.8

β

Let C be a bounded closed subset of the plane We claim there exists a square S that circumscribes C That is, C ⊆ S and each side of S touches C

Example 5.3.8

β

α

Let C be a bounded closed subset of the plane We claim there exists a square S that circumscribes C That is, C ⊆ S and each side of S touches C

Example 5.3.8

Now the circumscribing rectangle has become a square, with α < θ < β

θ β

Let C be a bounded closed subset of the plane We claim there exists a square S that circumscribes C That is, C ⊆ S and each side of S touches C

Example 5.3.8