Helly's selection theorem

Here is the proof from my lecture notes; I expect it is Helly's original proof. Today the theorem would perhaps be seen as an instance of weak$^*$ compactness.

Christer Bennewitz

Lemma(Helly). Suppose $\{\rho_j\}_1^\infty$ is a uniformly bounded sequence of increasing functions on an interval $I$. Then there is a subsequence converging pointwise to an increasing function.

Proof. Let $r_1,r_2,...$ be a dense sequence in $I$, for example an enumeration of the rational numbers in $I$. By Bolzano-Weierstrass’ theorem we may choose a subsequence $\{\rho_{1j}\}_{j=1}^\infty$ of $\{\rho_j\}_1^\infty$ so that $\rho_{1j}(r_1)$ converges. Similarly, we may choose a subsequence $\{\rho_{2j}\}_{j=1}^\infty$ of $\{\rho_{1j}\}_{j=1}^\infty$ such that $\rho_{2j}(r_2)$ converges; as a subsequence of $\{\rho_{1j}(r_1)\}_{j=1}^\infty$ also $\{\rho_{2j}(r_1)\}_{j=1}^\infty$ converges.

Continuing in this fashion, we obtain a sequence of sequences $\{\rho_{kj}\}_{j=1}^\infty$, $k=1,2,\dots$ such that each sequence is a subsequence of those preceding it, and such that $\rho(r_n)=\lim_{j\to\infty}\rho_{kj}(r_n)$ exists for $n\le k$. Thus $\rho_{jj}(r_n)\to\rho(r_n)$ as $j\to\infty$ for every $n$, since $\{\rho_{jj}(r_n)\}_1^\infty$ is a subsequence of $\{\rho_{nj}(r_n)\}_{j=1}^\infty$ from $j=n$ on. Clearly $\rho$ is increasing, so if $x\in I$ but $\ne r_n$ for all $n$, we may choose an increasing subsequence $\{r_{j_k}\}_1^\infty$ of $\{r_j\}_1^\infty$ converging to $x$, and define $\rho(x)=\lim_{k\to\infty}\rho(r_{j_k})$.

Suppose $x$ is a point of continuity of $\rho$. If $r_k<x<r_n$ we get $\rho_{jj}(r_k)−\rho(r_n)\le\rho_{jj}(x)−\rho(x)\le\rho_{jj}(r_n)−\rho(r_k)$. Given $\epsilon>0$ we may choose $k$ and $n$ such that $\rho(r_n)−\rho(r_k)<\epsilon$. We then obtain $$ −\epsilon\le\liminf_{j\to\infty}(\rho_{jj}(x)−\rho(x))\le\limsup_{j\to\infty} (ρ_{jj}(x)−\rho(x))\le\epsilon. $$ Hence $\{\rho_{jj}\}_1^\infty$ converges pointwise to $\rho$, except possibly in points of discontinuity of $\rho$. But there are at most countably many such discontinuities, $\rho$ being increasing. Hence repeating the trick of extracting subsequences, and then using the ‘diagonal’ sequence, we get a subsequence of the original sequence which converges everywhere in $I$.