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

Chapter 5Limits and Continuity... 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.. Definition 5.

Chapter 5

Limits and Continuity

Section 5.2 Continuous Functions

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

Definition 5.2.1

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

It follows that,

for any neighborhood V of f (c), f (U ∩ D) = { f (c)} ⊆ V Thus (c) always holds

Then there exists a neighborhood U of c such that U ∩ D = {c}

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)

Theorem 5.2.2

Proof:

Suppose first that c is an isolated point of D

Similarly, by Exercise 4.1.16, if (xn) is a sequence in D converging to c, then

xn ∈ U for all n greater than some M

But this implies that xn = c for n > M,

so limn → ∞ f (xn) = f (c) Thus (b) also holds

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

Then (a) ⇔ (d) is Definition 5.1.1,

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)

Theorem 5.2.2

Proof:

Now suppose that c is an accumulation point of D

(d) ⇔ (c) is Theorem 5.1.2, and (d) ⇔ (b) is essentially Theorem 5.1.8 ♦

Let’s look at this graphically

f (c)

f (c)+ε

f (c)–ε

f

c +

δ

c –

δ

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,

V

U

f is continuous at c iff

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.

Example 5.2.3

Let p be a polynomial We saw in Example 5.1.14 that for any c ∈ , lim x→c p(x) = p(c)

It follows that p is continuous on

Example 5.2.5

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.

given ε > 0, we may let δ = ε

Since for all x,

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

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).

Theorem 5.2.6

1

f x − f = x ≤ x

Example 5.2.8 describes a function (the Dirichlet function) that is discontinuous

at every real number

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 ∈

Define f : → by f (x) =

Example 5.2.9 describes a modified Dirichlet function

Define f : (0, 1)→ by f (x) =

•

1

2

1

3

1 2

1 3

2

f (1/2) = 1/2

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

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

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

1 , if is rational in lowest terms

0, if is irrational.

m n

x







f =

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

•

1

2

1

3

1 2

1 3

2

Let (xn) be a sequence of irrationals in (0, 1) with xn → c

Then f (xn) = 0 for all n, so lim f (xn) = 0.

But f (c) > 0 since c is rational.

So lim f (xn) ≠ f (c) and f is discontinuous at c

c

Example 5.2.9 describes a modified Dirichlet function

Define f : (0, 1)→ by f (x) =

1 , if is rational in lowest terms

0, if is irrational.

m n

x







1

2

1

3

1 2

1 3

2

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

There are only a finite number of rationals in (0,1) with denominators

less than k.

1/k

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

Given any ε > 0, there exists k ∈ such that 1/k < ε

d

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

Example 5.2.9 describes a modified Dirichlet function

Define f : (0, 1)→ by f (x) =

1 , if is rational in lowest terms

0, if is irrational.

m n

x







To show that f + g is continuous at c

it suffices by Theorem 5.2.2 to show that lim ( f + g)(xn) = ( f + g)(c)

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:

Theorem 5.2.10

Let (xn) be a sequence in D converging to c

From Definition 5.1.12 and Theorem 4.2.1, we have lim ( f + g)(xn) = lim [ f (xn) + g (xn)] = 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 (xn) so that g   (xn) ≠ 0 for all n Recall that the

quotient is defined only at those points x ∈ D for which g   (x) ≠ 0 ♦

D E

c

f (c)

g( f (c))

f

Let W be any neighborhood of (g ° f )(c) = g( f (c)).

W

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

a neighborhood V of f (c) such that

g( V ∩ E) ⊆ W

Since f is continuous at c, there exists

a neighborhood U of c such that

f ( U ∩ D) ⊆ V

Since f (D) ⊆ E, we have f(U ∩ D) ⊆ (V ∩ E)

If follows that g( f(U ∩ D)) ⊆ W,

so g ° f is continuous at c by Theorem 5.2.2

g

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:

f (c)

f

G

Suppose f is continuous Let G be an open set in We want to show f – 1(G) is open

Suppose c ∈ f – 1 (G) Then f (c) ∈ G.

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

V

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

⊆

Theorem 5.2.14

Corollary 5.2.15

A function f : D → is continuous on D iff for every open set G in there exists

an open set H in such that H ∩ D = f – 1(G).

A function f : → is continuous iff f – 1(G) is open in whenever G is open in

Proof:

f (c)

f

Conversely, suppose G being open implies f – 1(G) is open.

V

Given c ∈ , let V be a neighborhood of f (c). Then V is open

So, f – 1(V) is open Since f (c) ∈ V , c ∈ f – 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 ♦

f (U)

Theorem 5.2.14

Corollary 5.2.15

A function f : D → is continuous on D iff for every open set G in there exists

an open set H in such that H ∩ D = f – 1(G).

A function f : → is continuous iff f – 1(G) is open in whenever G is open in

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

Define f : → by f (x) =

x, if x ≤ 2,

4, if x > 2.

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

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

4

3

2

1

( ]

G

f – 1(G)

Example 5.2.16