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 $$S$$.

1. This will certainly be the case as soon as $$PQ/2^{n} < \epsilon$$.↩︎