71. Alternative proof of Weierstrass’s Theorem

The results of § 69 enable us to give an alternative proof of the important theorem proved in § 19.

If we divide \(PQ\) into two equal parts, one at least of them must contain infinitely many points of \(S\). We select the one which does, or, if both do, we select the left-hand half; and we denote the selected half by \(P_{1}Q_{1}\) (Fig. 28). If \(P_{1}Q_{1}\) is the left-hand half, \(P_{1}\) is the same point as \(P\).

Similarly, if we divide \(P_{1}Q_{1}\) into two halves, one at least of them must contain infinitely many points of \(S\). We select the half \(P_{2}Q_{2}\) which does so, or, if both do so, we select the left-hand half. Proceeding in this way we can define a sequence of intervals \[PQ,\quad P_{1}Q_{1},\quad P_{2}Q_{2},\quad P_{3}Q_{3},\ \dots,\] each of which is a half of its predecessor, and each of which contains infinitely many points of \(S\).

The points \(P\), \(P_{1}\), \(P_{2}\), … progress steadily from left to right, and so \(P_{n}\) tends to a limiting position \(T\). Similarly \(Q_{n}\) tends to a limiting position \(T’\). But \(TT’\) is plainly less than \(P_{n}Q_{n}\), whatever the value of \(n\); and \(P_{n}Q_{n}\), being equal to \(PQ/2^{n}\), tends to zero. Hence \(T’\) coincides with \(T\), and \(P_{n}\) and \(Q_{n}\) both tend to \(T\).

Then \(T\) is a point of accumulation of \(S\). For suppose that \(\xi\) is its coordinate, and consider any interval of the type \({[\xi – \epsilon, \xi + \epsilon]}\). If \(n\) is sufficiently large, \(P_{n}Q_{n}\) will lie entirely inside this interval.¹ Hence \({[\xi – \epsilon, \xi + \epsilon]}\) contains infinitely many points of \(S\).