100-104. Properties of continuous functions. Bounded functions. The oscillation of a function in an interval

100. The fundamental property of a continuous function.

It may perhaps be thought that the analysis of the idea of a continuous curve given in § 98 is not the simplest or most natural possible. Another method of analysing our idea of continuity is the following. Let $A$ and $B$ be two points on the graph of $ϕ (x)$ whose coordinates are $x_{0}$ , $ϕ (x_{0})$ and $x_{1}$ , $ϕ (x_{1})$ respectively. Draw any straight line $λ$ which passes between $A$ and $B$ . Then common sense certainly declares that if the graph of $ϕ (x)$ is continuous it must cut $λ$ .

If we consider this property as an intrinsic geometrical property of continuous curves it is clear that there is no real loss of generality in supposing $λ$ to be parallel to the axis of $x$ . In this case the ordinates of $A$ and $B$ cannot be equal: let us suppose, for definiteness, that $ϕ (x_{1}) > ϕ (x_{0})$ . And let $λ$ be the line $y = η$ , where $ϕ (x_{0}) < η < ϕ (x_{1})$ . Then to say that the graph of $ϕ (x)$ must cut $λ$ is the same thing as to say that there is a value of $x$ between $x_{0}$ and $x_{1}$ for which $ϕ (x) = η$ .

We conclude then that a continuous function $ϕ (x)$ must possess the following property: if $ϕ (x_{0}) = y_{0}, ϕ (x_{1}) = y_{1},$ and $y_{0} < η < y_{1}$ , then there is a value of $x$ between $x_{0}$ and $x_{1}$ for which $ϕ (x) = η$ . In other words as $x$ varies from $x_{0}$ to $x_{1}$ , $y$ must assume at least once every value between $y_{0}$ and $y_{1}$ .

We shall now prove that if $ϕ (x)$ is a continuous function of $x$ in the sense defined in § 98 then it does in fact possess this property. There is a certain range of values of $x$ , to the right of $x_{0}$ , for which $ϕ (x) < η$ . For $ϕ (x_{0}) < η$ , and so $ϕ (x)$ is certainly less than $η$ if $ϕ (x) - ϕ (x_{0})$ is numerically less than $η - ϕ (x_{0})$ . But since $ϕ (x)$ is continuous for $x = x_{0}$ , this condition is certainly satisfied if $x$ is near enough to $x_{0}$ . Similarly there is a certain range of values, to the left of $x_{1}$ , for which $ϕ (x) > η$ .

Let us divide the values of $x$ between $x_{0}$ and $x_{1}$ into two classes $L$ , $R$ as follows:

(1) in the class $L$ we put all values $ξ$ of $x$ such that $ϕ (x) < η$ when $x = ξ$ and for all values of $x$ between $x_{0}$ and $ξ$ ;

(2) in the class $R$ we put all the other values of $x$ , i.e. all numbers $ξ$ such that either $ϕ (ξ) \geq η$ or there is a value of $x$ between $x_{0}$ and $ξ$ for which $ϕ (x) \geq η$ .

Then it is evident that these two classes satisfy all the conditions imposed upon the classes $L$ , $R$ of § 17, and so constitute a section of the real numbers. Let $ξ_{0}$ be the number corresponding to the section.

First suppose $ϕ (ξ_{0}) > η$ , so that $ξ_{0}$ belongs to the upper class: and let $ϕ (ξ_{0}) = η + k$ , say. Then $ϕ (ξ^{'}) < η$ and so $ϕ (ξ_{0}) - ϕ (ξ^{'}) > k,$ for all values of $ξ^{'}$ less than $ξ_{0}$ , which contradicts the condition of continuity for $x = ξ_{0}$ .

Next suppose $ϕ (ξ_{0}) = η - k < η$ . Then, if $ξ^{'}$ is any number greater than $ξ_{0}$ , either $ϕ (ξ^{'}) \geq η$ or we can find a number $ξ ”$ between $ξ_{0}$ and $ξ^{'}$ such that $ϕ (ξ ”) \geq η$ . In either case we can find a number as near to $ξ_{0}$ as we please and such that the corresponding values of $ϕ (x)$ differ by more than $k$ . And this again contradicts the hypothesis that $ϕ (x)$ is continuous for $x = ξ_{0}$ .

Hence $ϕ (ξ_{0}) = η$ , and the theorem is established. It should be observed that we have proved more than is asserted explicitly in the theorem; we have proved in fact that $ξ_{0}$ is the least value of $x$ for which $ϕ (x) = η$ . It is not obvious, or indeed generally true, that there is a least among the values of $x$ for which a function assumes a given value, though this is true for continuous functions.

It is easy to see that the converse of the theorem just proved is not true. Thus such a function as the function $ϕ (x)$ whose graph is represented by Fig. 31 obviously assumes at least once every value between $ϕ (x_{0})$ and $ϕ (x_{1})$ : yet $ϕ (x)$ is discontinuous. Indeed it is not even true that $ϕ (x)$ must be continuous when it assumes each value once and once only. Thus let $ϕ (x)$ be defined as follows from $x = 0$ to $x = 1$ . If $x = 0$ let $ϕ (x) = 0$ ; if $0 < x < 1$ let $ϕ (x) = 1 - x$ ; and if $x = 1$ let $ϕ (x) = 1$ . The graph of the function is shown in Fig. 32; it includes the points $O$ , $C$ but not the points $A$ , $B$ . It is clear that, as $x$ varies from $0$ to $1$ , $ϕ (x)$ assumes once and once only every value between $ϕ (0) = 0$ and $ϕ (1) = 1$ ; but $ϕ (x)$ is discontinuous for $x = 0$ and $x = 1$ .

As a matter of fact, however, the curves which usually occur in elementary mathematics are composed of a finite number of pieces along which $y$ always varies in the same direction. It is easy to show that if $y = ϕ (x)$ always varies in the same direction, i.e. steadily increases or decreases, as $x$ varies from $x_{0}$ to $x_{1}$ , then the two notions of continuity are really equivalent, that if $ϕ (x)$ takes every value between $ϕ (x_{0})$ and $ϕ (x_{1})$ then it must be a continuous function in the sense of § 98. For let $ξ$ be any value of $x$ between $x_{0}$ and $x_{1}$ . As $x \to ξ$ through values less than $ξ$ , $ϕ (x)$ tends to the limit $ϕ (ξ - 0)$ (§ 95). Similarly as $x \to ξ$ through values greater than $ξ$ , $ϕ (x)$ tends to the limit $ϕ (ξ + 0)$ . The function will be continuous for $x = ξ$ if and only if $ϕ (ξ - 0) = ϕ (ξ) = ϕ (ξ + 0) .$ But if either of these equations is untrue, say the first, then it is evident that $ϕ (x)$ never assumes any value which lies between $ϕ (ξ - 0)$ and $ϕ (ξ)$ , which is contrary to our assumption. Thus $ϕ (x)$ must be continuous. The net result of this and the last section is consequently to show that our common-sense notion of what we mean by continuity is substantially accurate, and capable of precise statement in mathematical terms.

101.

In this and the following paragraphs we shall state and prove some general theorems concerning continuous functions.

Theorem 1. Suppose that $ϕ (x)$ is continuous for $x = ξ$ , and that $ϕ (ξ)$ is positive. Then we can determine a positive number $δ$ such that $ϕ (ξ)$ is positive throughout the interval $[ξ - δ, ξ + δ]$ .

For, taking $ϵ = \frac{1}{2} ϕ (ξ)$ in the fundamental inequality of § 98, we can choose $δ$ so that $| ϕ (x) - ϕ (ξ) | < \frac{1}{2} ϕ (ξ)$ throughout $[ξ - δ, ξ + δ]$ , and then $ϕ (x) \geq ϕ (ξ) - | ϕ (x) - ϕ (ξ) | > \frac{1}{2} ϕ (ξ) > 0,$ so that $ϕ (x)$ is positive. There is plainly a corresponding theorem referring to negative values of $ϕ (x)$ .

Theorem 2. If $ϕ (x)$ is continuous for $x = ξ$ , and $ϕ (x)$ vanishes for values of $x$ as near to $ξ$ as we please, or assumes, for values of $x$ as near to $ξ$ as we please, both positive and negative values, then $ϕ (ξ) = 0$ .

This is an obvious corollary of Theorem 1. If $ϕ (ξ)$ is not zero, it must be positive or negative; and if it were, for example, positive, it would be positive for all values of $x$ sufficiently near to $ξ$ , which contradicts the hypotheses of the theorem.

102. The range of values of a continuous function.

Let us consider a function $ϕ (x)$ about which we shall only assume at present that it is defined for every value of $x$ in an interval $[a, b]$ .

The values assumed by $ϕ (x)$ for values of $x$ in $[a, b]$ form an aggregate $S$ to which we can apply the arguments of § 80, as we applied them in § 81 to the aggregate of values of a function of $n$ . If there is a number $K$ such that $ϕ (x) \leq K$ , for all values of $x$ in question, we say that $ϕ (x)$ is bounded above. In this case $ϕ (x)$ possesses an upper bound $M$ : no value of $ϕ (x)$ exceeds $M$ , but any number less than $M$ is exceeded by at least one value of $ϕ (x)$ . Similarly we define ‘bounded below’, ‘lower bound’, ‘bounded’, as applied to functions of a continuous variable $x$ .

Theorem 1. If $ϕ (x)$ is continuous throughout $[a, b]$ , then it is bounded in $[a, b]$ .

We can certainly determine an interval $[a, ξ]$ , extending to the right from $a$ , in which $ϕ (x)$ is bounded. For since $ϕ (x)$ is continuous for $x = a$ , we can, given any positive number $ϵ$ however small, determine an interval $[a, ξ]$ throughout which $ϕ (x)$ lies between $ϕ (a) - ϵ$ and $ϕ (a) + ϵ$ ; and obviously $ϕ (x)$ is bounded in this interval.

Now divide the points $ξ$ of the interval $[a, b]$ into two classes $L$ , $R$ , putting $ξ$ in $L$ if $ϕ (ξ)$ is bounded in $[a, ξ]$ , and in $R$ if this is not the case. It follows from what precedes that $L$ certainly exists: what we propose to prove is that $R$ does not. Suppose that $R$ does exist, and let $β$ be the number corresponding to the section whose lower and upper classes are $L$ and $R$ . Since $ϕ (x)$ is continuous for $x = β$ , we can, however small $ϵ$ may be, determine an interval $[β - η, β + η]$ ¹ throughout which $ϕ (β) - ϵ < ϕ (x) < ϕ (β) + ϵ .$ Thus $ϕ (x)$ is bounded in $[β - η, β + η]$ . Now $β - η$ belongs to $L$ . Therefore $ϕ (x)$ is bounded in $[a, β - η]$ : and therefore it is bounded in the whole interval $[a, β + η]$ . But $β + η$ belongs to $R$ and so $ϕ (x)$ is not bounded in $[a, β + η]$ . This contradiction shows that $R$ does not exist. And so $ϕ (x)$ is bounded in the whole interval $[a, b]$ .

Theorem 2. If $ϕ (x)$ is continuous throughout $[a, b]$ , and $M$ and $m$ are its upper and lower bounds, then $ϕ (x)$ assumes the values $M$ and $m$ at least once each in the interval.

For, given any positive number $ϵ$ , we can find a value of $x$ for which $M - ϕ (x) < ϵ$ or $1 / {M - ϕ (x)} > 1 / ϵ$ . Hence $1 / {M - ϕ (x)}$ is not bounded, and therefore, by Theorem 1, is not continuous. But $M - ϕ (x)$ is a continuous function, and so $1 / {M - ϕ (x)}$ is continuous at any point at which its denominator does not vanish (Ex. XXXVII. 1). There must therefore be one point at which the denominator vanishes: at this point $ϕ (x) = M$ . Similarly it may be shown that there is a point at which $ϕ (x) = m$ .

The proof just given is somewhat subtle and indirect, and it may be well, in view of the great importance of the theorem, to indicate alternative lines of proof. It will however be convenient to postpone these for a moment.²

Example XXXVIII

1. If $ϕ (x) = 1 / x$ except when $x = 0$ , and $ϕ (x) = 0$ when $x = 0$ , then $ϕ (x)$ has neither an upper nor a lower bound in any interval which includes $x = 0$ in its interior, as e.g. the interval $[- 1, + 1]$ .

2. If $ϕ (x) = 1 / x^{2}$ except when $x = 0$ , and $ϕ (x) = 0$ when $x = 0$ , then $ϕ (x)$ has the lower bound $0$ , but no upper bound, in the interval $[- 1, + 1]$ .

3. Let $ϕ (x) = \sin (1 / x)$ except when $x = 0$ , and $ϕ (x) = 0$ when $x = 0$ . Then $ϕ (x)$ is discontinuous for $x = 0$ . In any interval $[- ϵ, + ϵ]$ the lower bound is $- 1$ and the upper bound $+ 1$ , and each of these values is assumed by $ϕ (x)$ an infinity of times.

4. Let $ϕ (x) = x - [x]$ . This function is discontinuous for all integral values of $x$ . In the interval $[0, 1]$ its lower bound is $0$ and its upper bound $1$ . It is equal to $0$ when $x = 0$ or $x = 1$ , but it is never equal to $1$ . Thus $ϕ (x)$ never assumes a value equal to its upper bound.

5. Let $ϕ (x) = 0$ when $x$ is irrational, and $ϕ (x) = q$ when $x$ is a rational fraction $p / q$ . Then $ϕ (x)$ has the lower bound $0$ , but no upper bound, in any interval $[a, b]$ . But if $ϕ (x) = (- 1)^{p} q$ when $x = p / q$ , then $ϕ (x)$ has neither an upper nor a lower bound in any interval.

103. The oscillation of a function in an interval.

Let $ϕ (x)$ be any function bounded throughout $[a, b]$ , and $M$ and $m$ its upper and lower bounds. We shall now use the notation $M (a, b)$ , $m (a, b)$ for $M$ , $m$ , in order to exhibit explicitly the dependence of $M$ and $m$ on $a$ and $b$ , and we shall write $O (a, b) = M (a, b) - m (a, b) .$

This number $O (a, b)$ , the difference between the upper and lower bounds of $ϕ (x)$ in $[a, b]$ , we shall call the oscillation of $ϕ (x)$ in $[a, b]$ . The simplest of the properties of the functions $M (a, b)$ , $m (a, b)$ , $O (a, b)$ are as follows.

(1) If $a \leq c \leq b$ then $M (a, b)$ is equal to the greater of $M (a, c)$ and $M (c, b)$ , and $m (a, b)$ to the lesser of $m (a, c)$ and $m (c, b)$ .

(2) $M (a, b)$ is an increasing, $m (a, b)$ a decreasing, and $O (a, b)$ an increasing function of $b$ .

(3) $O (a, b) \leq O (a, c) + O (c, b)$ .

The first two theorems are almost immediate consequences of our definitions. Let $μ$ be the greater of $M (a, c)$ and $M (c, b)$ , and let $ϵ$ be any positive number. Then $ϕ (x) \leq μ$ throughout $[a, c]$ and $[c, b]$ , and therefore throughout $[a, b]$ ; and $ϕ (x) > μ - ϵ$ somewhere in $[a, c]$ or in $[c, b]$ , and therefore somewhere in $[a, b]$ . Hence $M (a, b) = μ$ . The proposition concerning $m$ may be proved similarly. Thus (1) is proved, and (2) is an obvious corollary.

Suppose now that $M_{1}$ is the greater and $M_{2}$ the less of $M (a, c)$ and $M (c, b)$ , and that $m_{1}$ is the less and $m_{2}$ the greater of $m (a, c)$ and $m (c, b)$ . Then, since $c$ belongs to both intervals, $ϕ (c)$ is not greater than $M_{2}$ nor less than $m_{2}$ . Hence $M_{2} \geq m_{2}$ , whether these numbers correspond to the same one of the intervals $[a, c]$ and $[c, b]$ or not, and $O (a, b) = M_{1} - m_{1} \leq M_{1} + M_{2} - m_{1} - m_{2} .$ But $O (a, c) + O (c, b) = M_{1} + M_{2} - m_{1} - m_{2};$ and (3) follows.

104. Alternative proofs of Theorem 2 of § 102.

The most straightforward proof of Theorem 2 of § 102 is as follows. Let $ξ$ be any number of the interval $[a, b]$ . The function $M (a, ξ)$ increases steadily with $ξ$ and never exceeds $M$ . We can therefore construct a section of the numbers $ξ$ by putting $ξ$ in $L$ or in $R$ according as $M (a, ξ) < M$ or $M (a, ξ) = M$ . Let $β$ be the number corresponding to the section. If $a < β < b$ , we have $M (a, β - η) < M, M (a, β + η) = M$ for all positive values of $η$ , and so $M (β - η, β + η) = M,$ by (1) of § 103. Hence $ϕ (x)$ assumes, for values of $x$ as near as we please to $β$ , values as near as we please to $M$ , and so, since $ϕ (x)$ is continuous, $ϕ (β)$ must be equal to $M$ .

If $β = a$ then $M (a, a + η) = M$ . And if $β = b$ then $M (a, b - η) < M$ , and so $M (b - η, b) = M$ . In either case the argument may be completed as before.

The theorem may also be proved by the method of repeated bisection used in § 71. If $M$ is the upper bound of $ϕ (x)$ in an interval $P Q$ , and $P Q$ is divided into two equal parts, then it is possible to find a half $P_{1} Q_{1}$ in which the upper bound of $ϕ (x)$ is also $M$ . Proceeding as in § 71, we construct a sequence of intervals $P Q$ , $P_{1} Q_{1}$ , $P_{2} Q_{2}$ , … in each of which the upper bound of $ϕ (x)$ is $M$ . These intervals, as in § 71, converge to a point $T$ , and it is easily proved that the value of $ϕ (x)$ at this point is $M$ .

If $β = b$ we must replace this interval by $[β - η, β]$ , and $β + η$ by $β$ , throughout the argument which follows.↩︎
See § 104.↩︎

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98-99. Continuous functions of a real variable

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105-106. Sets of intervals on a line. The Heine-Borel Theorem

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