Continuity

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Chapter 4: Continuity Although we shall (in later chapters) be mainly interested in real and complex functions (i.e., in functions whose values are real or complex numbers) we shall also discuss vector-valued functions (i.e., functions with values in $\mathbb R^k$ ) and functions with values in an arbitrary metric space. The theorems we shall discuss in this general setting would not become any easier if we
restricted ourselves to real functions, for instance, and it actually simplifies and clarifies the picture to discard unnecessary hypotheses and to state and prove theorems in an appropriately general context.

The domains of definition of our functions will also be metric spaces, suitably specialized in various instances.

Contents
Contents
1.  Limits of functions
2.  Continuous functions
3.  Continuity and compactness
4.  Continuity and connectedness
5.  Discontinuities
6.  Monotonic functions
7.  Infinite limits and limits at infinity

1. Limits of functions

Definition 1. Let

X

and

Y

be metric spaces; suppose

E \subset X

f

maps

E

into

Y

, and

p

is a limit point of

E

. We write

f(x)\to q

x \to p

, or

\lim_{x\to p}f(x)=q

if there is a point

q \in Y

with the following property: For every

\varepsilon > 0

there exists a

\delta > 0

such that

d_Y(f(x),q)<\varepsilon

for all points

x\in E

for which

0<d_X(x,p)<\delta

The symbols $d_X$ and $d_Y$ refer to the distances in $X$ and $Y$ , respectively.

If $X$ and/or $Y$ are replaced by the real line, the complex plane, or by some Euclidean space $\mathbb R^k$ , the distances $d_X$ , $d_Y$ are of course replaced by absolute values, or by norms of differences. It should be noted that $p \in X$ , but that $p$ need not be a point of $E$ in the above definition. Moreover, even if $p\in E$ , we may very well have $f(p) \neq\lim_{x\to p}f(x)$ .

We can recast this definition in terms of limits of sequences:

Theorem 2.

\lim_{x\to p}f(x)=q

if and only if

\lim f(p_n)=q

for every sequence

\{p_n\}

E

such that

p_n\neq p

\lim p_n=p

Corollary 3. If

p

has a limit at

p

, this limit is unique.

Theorem 4. Suppose

E \subset X

, a metric space,

p

is a limit point of

E

f

and

g

are complex functions on

E

, and

\lim_{x\to p}f(x)=A

\lim_{x\to p}g(x)=B

. Then

\lim (f+g)(x)=A+B

;

\lim (fg)(x)=AB

;

\lim(f/g)(x)=A/B

, if

B\neq0

If $\mathbf f$ and $\mathbf g$ map $E$ into $\mathbb R^k$ , then $\lim(\mathbf{f}+\mathbf{g})(x)=\mathbf{A}+\mathbf{B}$ , $\lim(\mathbf{f}\cdot\mathbf{g})(x)=\mathbf{A}\cdot\mathbf{B}$ .

2. Continuous functions

Definition 5. Suppose

X

and

Y

are metric spaces,

E \subset X

p \in E

, and

f

maps

E

into

Y

. Then

f

is said to be continuous at

p

if for every

\varepsilon > 0

there exists a

\delta > 0

such that

d_Y(f(x),f(p))<\varepsilon

for all points

x\in E

for which

d_X(x,p)<\delta

If $f$ is continuous at every point of $E$ , then $f$ is said to be continuous on $E$ .

It should be noted that $f$ has to be defined at the point $p$ in order to be continuous at $p$ .

If $p$ is an isolated point of $E$ , then our definition implies that every function $f$ which has $E$ as its domain of definition is continuous at $p$ .

Theorem 6. In the situation given in Definition def42, assume also that

p

is a limit of

E

. Then

f

is continuous at

p

if and only if

\lim_{x\to p}f(x)=f(p)

We now turn to compositions of functions. A brief statement of the following theorem is that a continuous function of a continuous function is continuous.

Theorem 7. Suppose

X, Y, Z

are metric spaces,

E \subset X

f

maps

E

into

Y

g

maps the range of

f

f(E)

, into

Z

, and

h

is the mapping of

E

into

Z

defined by

h(x)=g(f(x))\, (x\in E)

. If

f

is continuous at a point

p\in E

and if

g

is continuous at the point

f(p)

, then

h

is continuous at

p

. This function

h

is called the composition or the composite of

f

and

g

Theorem 8. A mapping

f

of a metric space

X

into a metric space

Y

is continuous on

X

if and only if

f^{-1}(V)

is open in

X

for every open set

V

Y

. (This is a very useful characterization of continuity.)

Corollary 9. A mapping

f

of a metric space

X

into a metric space

Y

is continuous if and only if

f^{-1}(C)

is closed in

X

for every closed set

C

Y

Now turn to complex-valued and vector-valued functions, and to functions defined on subsets of $\mathbb R^k$ .

Theorem 10. Let

f

and

g

be complex continuous functions on a metric space

X

. Then

f + g

fg

, and

f/g

are continuous on

X

. (In the last case, we must of course assume that

g(x)\neq0

, for all

x\in X

Theorem 11. Let

f_1,\dots,f_k

be real functions on a metric space

X

, and let

\mathbf f

be the mapping of

X

into

\mathbb R^k

defined by

\mathbf f(x)=(f_1(x),\dots,f_k(x))\, (x\in X)

; then

f

is continuous if and only if each of the functions

f_1,\dots,f_k

is continuous;

If $\mathbf{f}$ and $\mathbf{g}$ are continuous mappings of $X$ into $\mathbb{R}^k$ , then $\mathbf{f} + \mathbf{g}$ and $\mathbf{f} \cdot \mathbf{g}$ are continuous on $X$ .

We defined the notion of continuity for functions defined on a subset $E$ of a metric space $X$ . However, the complement of $E$ in $X$ plays no role whatever in this definition (note that the situation was somewhat different for limits of functions). Accordingly, we lose nothing of interest by discarding the complement of the domain off This means that we may just as well talk only about continuous mappings of one metric space into another, rather than of mappings of subsets. This simplifies statements and proofs of some theorems.
3. Continuity and compactness

Theorem 12. Suppose

f

is a continuous mapping of a compact metric space

X

into a metric space

Y

. Then

f(X)

is compact.

We shall now deduce some consequences of Theorem theorem431.

Theorem 13. If

\mathbf f

is a continuous mapping of a compact metric space

X

into

\mathbb R^k

, then

f(X)

is closed and bounded. Thus,

f

is bounded.

The result is particularly important when $f$ is real:

Theorem 14.

f

is a continuous real function on a compact metric space

X

, and

M=\sup f(p)

m=\inf f(p)

. Then there exist points

p, q \in X

such that

f(p) = M

and

f(q) = m

The conclusion may also be stated as follows: There exist points $p$ and $q$ in $X$ such that $f(q)\leqslant f(x)\leqslant f(p)$ for all $x \in X$ ; that is, $f$ attains its maximum (at $p$ ) and its minimum (at $q$ ).

Theorem 15. Suppose

f

is a continuous

1

1

mapping of a compact metric space

X

onto a metric space

Y

. Then the inverse mapping

f^{-1}

defined on

Y

is a continuous mapping of

Y

onto

X

Definition 16. Let

f

be a mapping of a metric space

X

into a metric space

Y

. We say that

f

is {uniformly continuous} on

X

if for every

\varepsilon > 0

there exists

\delta> 0

such that

d_Y(f(p),f(q))<\varepsilon

for all

p

and

q

X

for which

d_X(p,q)<\delta

Let us consider the differences between the concepts of continuity and of uniform continuity. First, uniform continuity is a property of a function on a set, whereas continuity can be defined at a single point. To ask whether a given function is uniformly continuous at a certain point is meaningless. Second, if $f$ is continuous on $X$ , then it is possible to find, for each $\varepsilon > 0$ and for each point $p$ of $X$ , a number $\delta > 0$ having the property specified in Definition def42. This $\delta$ depends on $\varepsilon$ and on $p$ . If $f$ is, however, uniformly continuous on $X$ , then it is
possible, for each $\varepsilon > 0$ , to find one number $\delta> 0$ which will do for all points $p$ of $X$ .

Evidently, every uniformly continuous function is continuous. That the two concepts are equivalent on compact sets follows from the next theorem.

Theorem 17. Let

f

he a continuous mapping of a compact metric space

X

into a metric space

Y

. Then

f

is uniformly continuous on

X

Theorem 18. Let

E

be a noncompact set in

\mathbb R^1

. Then

there exists a continuous function on $E$ which is not bounded;
there exists a continuous and bounded function on $E$ which has no
maximum.

If, in addition,

E

is bounded, then:

there exists a continuous function on $E$ which is not uniformly continuous.

Proof. Theorem theorem431. Let $\{V_\alpha\}$ be an open cover of $f(X)$ . Since $f$ is continuous, each of the sets $f^{-1}(V_\alpha)$ is open. Since $X$ is compact, there are finitely many indices, say $\alpha_1,\dots,\alpha_n$ , such that $X\subset f^{-1}(V_{\alpha_1})\cup\dots\cup f^{-1}(V_{\alpha_n})$ . Since $f(f^{-1}(E))\subset E$ for every $E\subset Y$ , it implies that $f(X)\subset V_{\alpha_1}\cup\dots\cup V_{\alpha_n}$ .

Theorem theorem434. It suffices to prove that $f(V)$ is an open set in $Y$ for every open set $V$ in $X$ . Fix such a set $V$ . The complement $V^c$ of $V$ is closed in $X$ , hence compact; hence $f(V^c)$ is a compact subset of $Y$ and so is closed in $Y$ . Since $f$ is one-to-one and onto, $f(V)$ is the complement of $f(V^c)$ . Hence $f(V)$ is open.

Theorem theorem435. Let $\varepsilon > 0$ be given. Since $f$ is continuous, we can associate to each point $p \in X$ a positive number $\phi(p)$ such that
\begin{equation}
q\in X,\, d_X(p,q)<\phi(p)\Longrightarrow d_Y(f(p),f(q))<\frac{\varepsilon}{2}.
\end{equation}
Let $J(p)$ be the set of all $q \in X$ for which $d_X(p,q)<\frac12\phi(p)$ . Since $p\in J(p)$ , the collection of all sets $J(p)$ is an open cover of $X$ ; and since $X$ is compact, there is a finite set of points $p_1,\dots, p_n$ in $X$ , such that $X\subset J(p_1)\cup\dots\cup J(p_n)$ . We put $\delta=\frac12 \min[\phi(p_1),\dots,\phi(p_n)]>0$ . (This is one point where the finiteness of the covering, inherent in the definition of compactness, is essential. The minimum of a finite set of positive numbers is positive, whereas the inf of an infinite set of positive numbers may very well be 0.)

Now let $q$ and $p$ be points of $X$ , such that $d_X(p,q)<\delta$ . There is an integer $m$ , $1\leqslant m\leqslant n$ , such that $p\in J(p_m)$ ; hence $d_X(p,p_m)<\frac12\phi(p_m)$ , and we also have $d_X(q,p_m)\leqslant d_X(p,q)+d_X(p,p_m)<\delta+\frac12\phi(p_m)\leqslant \phi(p_m)$ . Finally, (41) shows that therefore $d_Y(f(p),f(q))\leqslant d_Y(f(p),f(p_m))+d_Y(f(q),f(p_m))<\varepsilon$ .

Theorem theorem436. Refer to the textbook.

We conclude this section by showing that compactness is also essential in Theorem theorem434.

{Example}
Let $X$ be the half-open interval $[0, 2\pi)$ on the real line, and let $f$ be the mapping of $X$ onto the circle $Y$ consisting of all points whose distance from the origin is 1, given by $\mathbf{f}(t)=(\cos t,\sin t)\, (0\leqslant t<2\pi)$ . $\mathbf{f}$ is a continuous 1-1 mapping of $X$ onto $Y$ .

However, the inverse mapping fails to be continuous at the point $(1, 0) = \mathbf{f}(0)$ . Of course, $X$ is not compact in this example. (It may be of interest to observe that $f^{-1}$ fails to be continuous in spite of the fact that $Y$ is compact!)
4. Continuity and connectedness

Theorem 19. If

f

is a continuous mapping of a metric space

X

into a metric space

Y

, and if

E

is a connected subset of

X

, then

f(E)

is connected.

Theorem 20. Let

f

be a continuous real function on the interval

[a, b]

. If

f(a)<f(b)

and if

c

is a number such that

f(a)<c<f(b)

, then there exists a point

x \in (a, b)

such that

f(x) =c

A similar result holds, of course, if $f(a) > f(b)$ . Roughly speaking, the theorem says that a continuous real function assumes all intermediate values on an interval.
5. Discontinuities If $x$ is a point in the domain of definition of the function $f$ at which $f$ is not continuous, we say that $f$ is discontinuous at $x$ , or that $f$ has a discontinuity at $x$ . If $f$ is defined on an interval or on a segment, it is customary to divide discontinuities into two types.

{Example}
Define $f(x)=\sin \frac1x\, (x\neq0)$ , and $f(0)=0$ . Since neither $f(0+)$ nor $f(0-)$ exists, $f$ has a discontinuity of the
second kind at $x = 0$ . $f$ is continuous at every point $x\neq0$ .

It might seem that Theorem theorem442 has a converse. That this is not so may be concluded from this example.
6. Monotonic functions We shall now study those functions which never decrease (or never increase) on a given segment.

Theorem 21. Let

f

be monotonically increasing on

(a, b)

. Then

f(x+)

and

f(x-)

exist at every point of

x

(a, b)

. More precisely,
\[\sup_{a<t<x}f(t)=f(x-)\leqslant f(x)\leqslant f(x+)=\inf_{x<t<b}f(t).\]

Furthermore, if $a<x<y<b$ , then $f(x+)\leqslant f(y-)$ .

Analogous results evidently hold for monotonically decreasing functions.

Theorem 22. Monotonic functions have no discontinuities of the second kind.

This corollary implies that every monotonic function is discontinuous at a countable set of points at most. Instead of appealing to the general theorem whose proof is sketched in Theorem theorem464, we give here a simple proof which is applicable to monotonic functions.

Theorem 23. Let

f

be monotonic on

(a, b)

. Then the set of points of

(a, b)

at which

f

is discontinuous is at most countable.

Theorem 24. Let

f

be a real function defined on

(a, b)

. The set of points at which

f

has a simple discontinuity is at most countable.

Proof. Let $E$ be the set on which $f(x-) <f(x+)$ . With each point $x$ of $E$ , associate a triple $(p, q, r)$ of rational
numbers such that

$f(x-)<p<f(x+)$ ,
$a<q<t<x$ implies $f(t)<p$ ,
$x<t<r<b$ implies $f(t)>p$ .

The set of all such triples is countable. It can be shown that each triple is associated with at most one point of $E$ . Deal similarly with the other possible types of simple discontinuities, then the proof is completed.

{Remark}
It should be noted that the discontinuities of a monotonic function need not be isolated. In fact, given any countable subset $E$ of $(a, b)$ , which may even be dense, we can construct a function $f$ , monotonic on $(a, b)$ , discontinuous at every point of $E$ , and at no other point of $(a, b)$ .

To show this, let the points of $E$ be arranged in a sequence $\{x_n\},\, n = 1, 2, 3,\dots$ . Let $\{c_n\}$ be a sequence of positive numbers such that $\sum c_n$ converges. Define $f(x)=\sum_{x_n<x}c_n\, (a<x<b)$ . Since it converges absolutely, the order in which the terms are arranged is immaterial. $f$ has the following properties:

$f$ is monotonically increasing on $(a, b)$ ;
$f$ is discontinuous at every point of $E$ ; in fact, $f(x_n+)-f(x_n-)=c_n$ ,
$f$ is continuous at every other point of $(a, b)$ .

Moreover, it is not hard to see that $f(x-) =f(x)$ at all points of $(a, b)$ . If a function satisfies this condition, we say that $f$ is continuous from the left. If the summation in $f$ were taken over all indices $n$ for which $x_n \leqslant x$ , we would have $f(x+) = f(x)$ at every point of $(a, b)$ ; that is, $f$ would be continuous from the right.

Functions of this sort can also be defined by another method.
7. Infinite limits and limits at infinity To enable us to operate in the extended real number system, we shall now enlarge the scope of the definition of limit of function, by reformulating it in terms of neighborhoods.

For any real number $x$ , we have already defined a neighborhood of $x$ to be any segment $(x-\delta,x+\delta)$ .

Definition 25. Let

f

be a real function defined on

E \subset \mathbb{R}

. We say that

f(t)\to A

t\to x

, where

A

and

x

are in the extended real number system, if for every neighborhood

U

A

there is a neighborhood

V

x

such that

V \cap E

is not empty, and such that

f(t) \in U

for all

t\in V\cap E,\, t\neq x

A moment's consideration will show that this coincides with the original definition when $A$ and $x$ are real.

The analogue of Theorem theorem424 is still true, and the proof offers nothing new.

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Continuity

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