Analysis with an introduction to proof 5th by steven lay ch05

Chapter 5
Limits and Continuity

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 1


Section 5.2
Continuous Functions

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Section 5.2, Slide 2


Definition 5.2.1
Let f : D →

and let c ∈ D. We say that f is continuous at c if
for each ε > 0 there exists a δ > 0 such that | f (x) – f (c)| < ε
whenever | x – c | < δ and x ∈ D .

If f is continuous at each point of a subset S of D, then f is said to be continuous on S.

Notice that the definition of continuity at a point c requires c to be in D, but it
does not require c to be an accumulation point of D.
In fact, if c is an isolated point of D, then f is automatically continuous at c.
For if c is an isolated point of D, then there exists a δ > 0 such that, if | x – c | < δ
and x ∈ D, then x = c.Thus whenever | x – c | < δ and x ∈ D, we have
| f (x) – f (c) | = 0 < ε for all ε > 0. Hence f is continuous at c.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 3


Theorem 5.2.2
Let f : D →

and let c ∈ D. The following three conditions are equivalent:

(a) f is continuous at c.
(b) If (xn) is any sequence in D such that (xn) converges to c, then
lim n → ∞ f (xn) = f (c).
(c) For every neighborhood V of f (c) there exists a neighborhood U of c
such that f (U ∩ D) ⊆ V.
And, if c is an accumulation point of D, then the above are all equivalent to
(d) f has a limit at c and lim x → c f (x) = f (c).
Proof:
Suppose first that c is an isolated point of D.
Then there exists a neighborhood U of c such that U ∩ D = {c}.It follows that,
for any neighborhood V of f (c), f (U ∩ D) = { f (c)} ⊆ V. Thus (c) always holds.
Similarly, by Exercise 4.1.16, if (x ) is a sequence in D converging to c, then
n
x ∈ U for all n greater than some M. But this implies that xn = c for n > M,
n
so lim
f (x ) = f (c). Thus (b) also holds.
n → ∞
n

We have already observed that (a) applies, so (a), (b), and (c) are all equivalent.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 4


Theorem 5.2.2
Let f : D →

and let c ∈ D. The following three conditions are equivalent:

(a) f is continuous at c.
(b) If (xn) is any sequence in D such that (xn) converges to c, then
lim n → ∞ f (xn) = f (c).
(c) For every neighborhood V of f (c) there exists a neighborhood U of c
such that f (U ∩ D) ⊆ V.
And, if c is an accumulation point of D, then the above are all equivalent to
(d) f has a limit at c and lim x → c f (x) = f (c).
Proof:
Now suppose that c is an accumulation point of D.
Then (a) ⇔ (d) is Definition 5.1.1,
(d) ⇔ (c) is Theorem 5.1.2,
and (d) ⇔ (b) is essentially Theorem 5.1.8. ♦

Let’s look at this graphically.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 5

f is continuous at c iff
for each ε > 0

there exists a δ > 0
such that | f (x) – f (c)| < ε whenever | x – c | < δ and x ∈ D.
Or, in terms of neighborhoods, for each neighborhood V of f (c),
such that f (U ∩ D) ⊆ V.

there exists a neighborhood U of c,

f

f (c) + ε

f (c)

V
f (c) – ε

c–

δ

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

c
U

c+

δ

Section 5.2, Slide 6


Example 5.2.3
Let p be a polynomial. We saw in Example 5.1.14 that for any c ∈
It follows that p is continuous on

, lim x → c p(x) = p(c).

.

Example 5.2.5
Let f (x) = x sin (1/x) for x ≠ 0 and f (0) = 0. Here is the graph:

It appears that f may be
continuous at x = 0.
Let’s prove that it is.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 7


We claim that the function defined by f (x) = x sin (1/x) for x ≠ 0 and f (0) = 0
is continuous at x = 0.

f ( Since
x) − f (0) = x sin

1
≤for|allx x,|
x

given ε > 0, we may let δ = ε .

Then when | x – 0 | < δ we have | f (x) – f (0)| ≤ | x | < δ = ε .

The negation of (a) and (b) in Theorem 5.2.2 gives a useful test of discontinuity.

Theorem 5.2.6
Let f : D →

and let c ∈ D. Then f is discontinuous at c iff there exists a

sequence (xn) in D such that (xn) converges to c but the sequence ( f (xn)) does
not converge to f (c).

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 8


Example 5.2.8 describes a function (the Dirichlet function) that is discontinuous
at every real number.

Define f :

→

by

f (x) =

1,

if x is rational,

0,

if x is irrational.

If c ∈

, then every neighborhood of c will

contain rational points at which f (x) = 1 and
also irrational points at which f (x) = 0.

Thus lim x → c f (x) cannot possibly exist and f is discontinuous at each c ∈

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

.

Section 5.2, Slide 9

Example 5.2.9 describes a modified Dirichlet function.

Define f : (0, 1) →

1
n ,

0,

by f (x) =

1

if x =

m
is rational in lowest terms
n

if x is irrational.
f (1/2) = 1/2

•

2

1

•

f (1/3) = f (2/3) = 1/3

•

3

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f (1/4) = f (3/4) = 1/4

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1

1

2

3

2

3

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f

••
•

(

)

1/ 2 = 0

1

We claim that f is continuous at each irrational number in (0, 1) and
discontinuous at each rational in (0, 1).

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 10


Example 5.2.9 describes a modified Dirichlet function.

Define f : (0, 1) →

1
n ,

0,

by f (x) =

1

if x =

if x is irrational.

•

Let c be a rational number in (0, 1).

2

1

•

Let (xn) be a sequence of irrationals in

•

3

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(0, 1) with xn → c.

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1
3

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c

•
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•

m
is rational in lowest terms
n

•
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1

2

2

3

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Then f (xn) = 0 for all n, so
•

•
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lim f (xn) = 0.

1

But f (c) > 0 since c is rational.
So lim f (xn) ≠ f (c) and f is discontinuous at c.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 11

Example 5.2.9 describes a modified Dirichlet function.

Define f : (0, 1) →

1
n ,

0,

by f (x) =

1

if x =

m
is rational in lowest terms
n

if x is irrational.

•

Let d be an irrational number in (0, 1).

2

1

•

Given any ε > 0, there exists k ∈

•

such that 1/k < ε .

3

•

1/k
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There are only a finite number of

••
•

rationals in (0, 1) with denominators
less than k.

1

d

3

1

2

2

3

1

Thus there exists a δ > 0 such that all the rationals in (d – δ, d + δ ) have a
denominator (in lowest terms) greater than or equal to k.
It follows that if x ∈ (0, 1) and | x – d | < δ , then | f (x) – f (d)| = | f (x)| ≤ 1/k < ε .
Hence f is continuous at d.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 12


Theorem 5.2.10
Let f and g be functions form D to

and let c ∈ D. Suppose that f and g

are continuous at c. Then
(a) f + g and f g are continuous at c, and
(b) f /g is continuous at c if g(c) ≠ 0.
Proof:
Let (x ) be a sequence in D converging to c.
n

To show that f + g is continuous at c

it suffices by Theorem 5.2.2 to show that lim (  f + g)(x ) = ( f + g)(c).
n
From Definition 5.1.12 and Theorem 4.2.1, we have
lim ( f + g)(x ) = lim [ f (xn) + g (xn)]
n

= lim  f (xn) + lim g (xn)
=  f (c) + g(c) = ( f + g)(c).

The proofs for the product and quotient are similar, the only difference being that
for f /g we have to choose the sequence (x ) so that g (x ) ≠ 0 for all n. Recall that the
n
n
quotient is defined only at those points x ∈ D for which g (x) ≠ 0. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 13


Theorem 5.2.12
Let f : D →

and g: E →

be functions such that f (D) ⊆ E. If f is

continuous at a point c ∈ D and g is continuous at f (c), then the composition

g°f:D→

is continuous at c.

Proof:
Let W be any neighborhood of

D

E
( g ° f )(c) = g( f (c)).
W

U

Since g is continuous at f (c), there exists

V

a neighborhood V of f (c) such that

g( f (c))

c
f

f (c)

g

g(V ∩ E)

⊆

W.

Since f is continuous at c, there exists
a neighborhood U of c such that
f (U ∩ D)

Since f (D)

⊆

E, we have f (U ∩ D)

If follows that g( f (U ∩ D))

⊆

⊆

⊆

V.

(V ∩ E).

W,

so g ° f is continuous at c by Theorem 5.2.2.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 14


Theorem 5.2.14
A function f : D →
an open set H in

is continuous on D iff for every open set G in
such that H ∩ D = f

–1

there exists

(G).

Corollary 5.2.15
A function f :

→

is continuous iff f

–1

(G) is open in

whenever G is open in

.

f
f

–1

(G ）

U

G

V
c

f (c)

Proof:
Suppose f is continuous. Let G be an open set in
Suppose c ∈ f

–1

.

We want to show f

–1

(G) is open.

Then f (c) ∈ G.

(G).

Since G is open, there exists a neighborhood V of f (c) such that V

⊆ G.

Since f is continuous, there exists a neighborhood U of c such that f (U)
But then U ⊆ f

–1

(G) so c is an interior point of f

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

–1

(G) and f

–1

⊆

V

⊆ G.

(G) is open.

Section 5.2, Slide 15


Theorem 5.2.14
A function f : D →
an open set H in

is continuous on D iff for every open set G in
such that H ∩ D = f

–1

there exists

(G).

Corollary 5.2.15
A function f :

→

is continuous iff f

–1

(G) is open in

whenever G is open in

.

f
f

–1

(V ）

U

V
c

f (c)

Conversely, suppose G being open implies f
Given c ∈
So, f

–1

(V) is open.

f (U)

–1

(G) is open.

, let V be a neighborhood of f (c).
Since f (c) ∈ V, c ∈ f

Then V is open.
–1

(V).

Thus there exists a neighborhood U of c such that U ⊆ f

–1

(V).

But then f (U ) ⊆ V , and f is continuous at c. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.2, Slide 16


The following example shows how a discontinuous function might fail to have
the pre-image of an open set be open.

Example 5.2.16

Define f :

→

if x ≤ 2,

4,

if x > 2.

by f (x) =

Let G be the open interval (1, 3).

4

(

3

x,

)

Then f

G

–1

(G) = (1, 2], which is not open.

2

1

(

]

1

2

f

–1

3

4

(G)

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Section 5.2, Slide 17