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

We shall now proceed to prove some theorems concerning the oscillation of a function which are of a somewhat abstract character but of very great importance, particularly, as we shall see later, in the theory of integration. These theorems depend upon a general theorem concerning intervals on a line.

Suppose that we are given a set of intervals in a straight line, that is to say an aggregate each of whose members is an interval $${[\alpha, \beta]}$$. We make no restriction as to the nature of these intervals; they may be finite or infinite in number; they may or may not overlap;1 and any number of them may be included in others.

It is worth while in passing to give a few examples of sets of intervals to which we shall have occasion to return later.

(i) If the interval $${[0, 1]}$$ is divided into $$n$$ equal parts then the $$n$$ intervals thus formed define a finite set of non-overlapping intervals which just cover up the line.

(ii) We take every point $$\xi$$ of the interval $${[0, 1]}$$, and associate with $$\xi$$ the interval $${[\xi – \delta, \xi + \delta]}$$, where $$\delta$$ is a positive number less than $$1$$, except that with $$0$$ we associate $${[0, \delta]}$$ and with $$1$$ we associate $${[1 – \delta, 1]}$$, and in general we reject any part of any interval which projects outside the interval $${[0, 1]}$$. We thus define an infinite set of intervals, and it is obvious that many of them overlap with one another.

(iii) We take the rational points $$p/q$$ of the interval $${[0, 1]}$$, and associate with $$p/q$$ the interval ${\left[\frac{p}{q} – \frac{\delta}{q^{3}}, \frac{p}{q} + \frac{\delta}{q^{3}}\right]},$ where $$\delta$$ is positive and less than $$1$$. We regard $$0$$ as $$0/1$$ and $$1$$ as $$1/1$$: in these two cases we reject the part of the interval which lies outside $${[0, 1]}$$. We obtain thus an infinite set of intervals, which plainly overlap with one another, since there are an infinity of rational points, other than $$p/q$$, in the interval associated with $$p/q$$.

The Heine-Borel Theorem. Suppose that we are given an interval $${[a, b]}$$, and a set of intervals $$I$$ each of whose members is included in $${[a, b]}$$. Suppose further that $$I$$ possesses the following properties:

(i) every point of $${[a, b]}$$, other than $$a$$ and $$b$$, lies inside2 at least one interval of $$I$$;

(ii) $$a$$ is the left-hand end point, and $$b$$ the right-hand end point, of at least one interval of $$I$$.

Then it is possible to choose a finite number of intervals from the set $$I$$ which form a set of intervals possessing the properties (i) and (ii).

We know that $$a$$ is the left-hand end point of at least one interval of $$I$$, say $${[a, a_{1}]}$$. We know also that $$a_{1}$$ lies inside at least one interval of $$I$$, say $${[a_{1}’, a_{2}]}$$. Similarly $$a_{2}$$ lies inside an interval $${[a_{2}’, a_{3}]}$$ of $$I$$. It is plain that this argument may be repeated indefinitely, unless after a finite number of steps $$a_{n}$$ coincides with $$b$$.

If $$a_{n}$$ does coincide with $$b$$ after a finite number of steps then there is nothing further to prove, for we have obtained a finite set of intervals, selected from the intervals of $$I$$, and possessing the properties required. If $$a_{n}$$ never coincides with $$b$$, then the points $$a_{1}$$, $$a_{2}$$, $$a_{3}$$, … must (since each lies to the right of its predecessor) tend to a limiting position, but this limiting position may, so far as we can tell, lie anywhere in $${[a, b]}$$.

Let us suppose now that the process just indicated, starting from $$a$$, is performed in all possible ways, so that we obtain all possible sequences of the type $$a_{1}$$, $$a_{2}$$, $$a_{3}$$, …. Then we can prove that there must be at least one such sequence which arrives at $$b$$ after a finite number of steps.

There are two possibilities with regard to any point $$\xi$$ between $$a$$ and $$b$$. Either (i) $$\xi$$ lies to the left of some point $$a_{n}$$ of some sequence or (ii) it does not. We divide the points $$\xi$$ into two classes $$L$$ and $$R$$ according as to whether (i) or (ii) is true. The class $$L$$ certainly exists, since all points of the interval $${[a, a_{1}]}$$ belong to $$L$$. We shall now prove that $$R$$ does not exist, so that every point $$\xi$$ belongs to $$L$$.

If $$R$$ exists then $$L$$ lies entirely to the left of $$R$$, and the classes $$L$$, $$R$$ form a section of the real numbers between $$a$$ and $$b$$, to which corresponds a number $$\xi_{0}$$. The point $$\xi_{0}$$ lies inside an interval of $$I$$, say $${[\xi’, \xi”]}$$, and $$\xi’$$ belongs to $$L$$, and so lies to the left of some term $$a_{n}$$ of some sequence. But then we can take $${[\xi’, \xi”]}$$ as the interval $${[a_{n}’, a_{n+1}]}$$ associated with $$a_{n}$$ in our construction of the sequence $$a_{1}$$, $$a_{2}$$, $$a_{3}$$, …; and all points to the left of $$\xi”$$ lie to the left of $$a_{n+1}$$. There are therefore points of $$L$$ to the right of $$\xi_{0}$$, and this contradicts the definition of $$R$$. It is therefore impossible that $$R$$ should exist.

Thus every point $$\xi$$ belongs to $$L$$. Now $$b$$ is the right-hand end point of an interval of $$I$$, say $${[b_{1}, b]}$$, and $$b_{1}$$ belongs to $$L$$. Hence there is a member $$a_{n}$$ of a sequence $$a_{1}$$, $$a_{2}$$, $$a_{3}$$, … such that $$a_{n} > b_{1}$$. But then we may take the interval $${[a_{n}’, a_{n+1}]}$$ corresponding to $$a_{n}$$ to be $${[b_{1}, b]}$$, and so we obtain a sequence in which the term after the $$n$$th coincides with $$b$$, and therefore a finite set of intervals having the properties required. Thus the theorem is proved.

It is instructive to consider the examples of § 105 in the light of this theorem.

(i) Here the conditions of the theorem are not satisfied: the points $$1/n$$, $$2/n$$, $$3/n$$, … do not lie inside any interval of $$I$$.

(ii) Here the conditions of the theorem are satisfied. The set of intervals ${[0, 2\delta]}, \quad {[\delta, 3\delta]}, \quad {[2\delta, 4\delta]}, \ \dots, \quad {[1 – 2\delta, 1]},$ associated with the points $$\delta$$, $$2\delta$$, $$3\delta$$, …, $$1 – \delta$$, possesses the properties required.

(iii) In this case we can prove, by using the theorem, that there are, if $$\delta$$ is small enough, points of $${[0, 1]}$$ which do not lie in any interval of $$I$$.

If every point of $${[0, 1]}$$ lay inside an interval of $$I$$ (with the obvious reservation as to the end points), then we could find a finite number of intervals of $$I$$ possessing the same property and having therefore a total length greater than $$1$$. Now there are two intervals, of total length $$2\delta$$, for which $$q = 1$$, and $$q – 1$$ intervals, of total length $$2\delta(q – 1)/q^{3}$$, associated with any other value of $$q$$. The sum of any finite number of intervals of $$I$$ can therefore not be greater than $$2\delta$$ times that of the series $1 + \frac{1}{2^{3}} + \frac{2}{3^{3}} + \frac{3}{4^{3}} + \dots,$ which will be shown to be convergent in Ch. VIII. Hence it follows that, if $$\delta$$ is small enough, the supposition that every point of $${[0, 1]}$$ lies inside an interval of $$I$$ leads to a contradiction.

The reader may be tempted to think that this proof is needlessly elaborate, and that the existence of points of the interval, not in any interval of $$I$$, follows at once from the fact that the sum of all these intervals is less than $$1$$. But the theorem to which he would be appealing is (when the set of intervals is infinite) far from obvious, and can only be proved rigorously by some such use of the Heine-Borel Theorem as is made in the text.

## 106.

We shall now apply the Heine-Borel Theorem to the proof of two important theorems concerning the oscillation of a continuous function.

Theorem I. If $$\phi(x)$$ is continuous throughout the interval $${[a, b]}$$, then we can divide $${[a, b]}$$ into a finite number of sub-intervals $${[a, x_{1}]}$$, $${[x_{1}, x_{2}]}$$, …, $${[x_{n}, b]}$$, in each of which the oscillation of $$\phi(x)$$ is less than an assigned positive number $$\epsilon$$.

Let $$\xi$$ be any number between $$a$$ and $$b$$. Since $$\phi(x)$$ is continuous for $$x = \xi$$, we can determine an interval $${[\xi – \delta, \xi + \delta]}$$ such that the oscillation of $$\phi(x)$$ in this interval is less than $$\epsilon$$. It is indeed obvious that there are an infinity of such intervals corresponding to every $$\xi$$ and every $$\epsilon$$, for if the condition is satisfied for any particular value of $$\delta$$, then it is satisfied a fortiori for any smaller value. What values of $$\delta$$ are admissible will naturally depend upon $$\xi$$; we have at present no reason for supposing that a value of $$\delta$$ admissible for one value of $$\xi$$ will be admissible for another. We shall call the intervals thus associated with $$\xi$$ the $$\epsilon$$-intervals of $$\xi$$.

If $$\xi = a$$ then we can determine an interval $${[a, a + \delta]}$$, and so an infinity of such intervals, having the same property. These we call the $$\epsilon$$-intervals of $$a$$, and we can define in a similar manner the $$\epsilon$$-intervals of $$b$$.

Consider now the set $$I$$ of intervals formed by taking all the $$\epsilon$$-intervals of all points of $${[a, b]}$$. It is plain that this set satisfies the conditions of the Heine-Borel Theorem; every point interior to the interval is interior to at least one interval of $$I$$, and $$a$$ and $$b$$ are end points of at least one such interval. We can therefore determine a set $$I’$$ which is formed by a finite number of intervals of $$I$$, and which possesses the same property as $$I$$ itself.

The intervals which compose the set $$I’$$ will in general overlap as in Fig. 34. But their end points obviously divide up $${[a, b]}$$ into a finite set of intervals $$I”$$ each of which is included in an interval of $$I’$$, and in each of which the oscillation of $$\phi(x)$$ is less than $$\epsilon$$. Thus Theorem I is proved.

Theorem II. Given any positive number $$\epsilon$$, we can find a number $$\eta$$ such that, if the interval $${[a, b]}$$ is divided in any manner into sub-intervals of length less than $$\eta$$, then the oscillation of $$\phi(x)$$ in each of them will be less than $$\epsilon$$.

Take $$\epsilon_{1} < \frac{1}{2}\epsilon$$, and construct, as in Theorem I, a finite set of sub-intervals $$j$$ in each of which the oscillation of $$\phi(x)$$ is less than $$\epsilon_{1}$$. Let $$\eta$$ be the length of the least of these sub-intervals $$j$$. If now we divide $${[a, b]}$$ into parts each of length less than $$\eta$$, then any such part must lie entirely within at most two successive sub-intervals $$j$$. Hence, in virtue of (3) of § 103, the oscillation of $$\phi(x)$$, in one of the parts of length less than $$\eta$$, cannot exceed twice the greatest oscillation of $$\phi(x)$$ in a sub-interval $$j$$, and is therefore less than $$2\epsilon_{1}$$, and therefore than $$\epsilon$$.

This theorem is of fundamental importance in the theory of definite integrals (Ch. VII). It is impossible, without the use of this or some similar theorem, to prove that a function continuous throughout an interval necessarily possesses an integral over that interval.

1. The word overlap is used in its obvious sense: two intervals overlap if they have points in common which are not end points of either. Thus $${[0, \frac{2}{3}]}$$ and $${[\frac{1}{3}, 1]}$$ overlap. A pair of intervals such as $${[0, \frac{1}{2}]}$$ and $${[\frac{1}{2}, 1]}$$ may be said to abut.↩︎
2. That is to say ‘in and not at an end of’.↩︎