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 $$PQ,P_1{Q}_1,P_2{Q}_2,P_3{Q}_3,\dots ,$$ 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 $T^{\prime }$. But $T{T}^{\prime }$ 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 $T^{\prime }$ 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 $[\xi –ϵ,\xi +ϵ]$. If $n$ is sufficiently large, $P_n{Q}_n$ will lie entirely inside this interval.¹ Hence $[\xi –ϵ,\xi +ϵ]$ contains infinitely many points of $S$.