# An Analysis of the First Proofs of the Heine-Borel Theorem - Schoenflies' Proof

Author(s):
Nicole R. Andre (Wittenberg University), Susannah M. Engdahl (Wittenberg University), and Adam E. Parker (Wittenberg University)

### Schoenflies' Proof

The next proof is due to Arthur Schoenflies in an 1899 review of point-set topology that he wrote for the German Mathematical Association [15]. In practice, this proof is very similar to that of Borel, though it contains more details. Schoenflies assumed the same monotone convergence version of completeness as Borel. Before stating the theorem, he attached both Borel's and Heine’s names to it:

To give a last example, I prove the following theorem of Borel’s, which extends a known theorem of Heine:

V. If on a straight line there is an infinite sequence of intervals $\delta,$ so that every point of the interval $a \dots b$ is an interior point of at least one interval $\delta,$ then there is also always a finite subset of such intervals.

In the following passage, Schoenflies chose an arbitrary point  $a_1$ in the interval and took $\delta_1$ to be any interval containing it. He looked at the left endpoint $a_2$ of $\delta_1$ and let $\delta_2$ be any interval containing $a_2.$ Continuing to the left in this fashion, he obtained a sequence of points $a_1,$ $a_2,$ … . Assuming that $a$ is not reached in a finite number of steps (else $\left[a, a_1\right]$ is covered by a finite number of intervals and we can proceed by looking to the right of $a_1$), then the sequence of left endpoints is infinite, decreasing, and bounded below by $a.$ By either the monotone convergence or Bolzano-Weierstrass properties, it must have a limit point, which he called $a_{\omega}.$

Let $a_1$ be an arbitrary point and $\delta_1$ an accompanying interval and, further, let $a_2$ be the left endpoint of $\delta_1$ and ${\delta}_2$ its accompanying interval. Likewise let $a_3$ be the left endpoint of ${\delta}_2$ etc. If $a$ is not yet arrived at by means of a finite number of intervals $\delta_1,$ ${\delta}_2$... ${\delta}_{\gamma},$ then the points

$a_1$, $a_2$, $a_3$…$a_{\gamma}$

have a limit $a_{\omega}$ which belongs to an interval ${\delta}_{\omega}.$

Schoenflies continued by applying the same technique to $a_{\omega}$ and ${\delta}_{\omega},$ and obtained an infinite sequence of points

$a_1$, $a_2$, …, $a_{\omega},$ …, $a_{\alpha},$ …

which he said was countable by a previous theorem.

Then let $a_{\omega+1}$ be the left endpoint of ${\delta}_{\omega},$ ${\delta}_{\omega +1}$ the accompanying interval and $a_{\omega+2}$ its left endpoint etc. We then arrive at a well defined sequence of points ...

$a_1$, $a_2$,…$a_{\omega},$…$a_{\alpha},$…

respectively of intervals...

$\delta_1,$ $\delta_2,$…${\delta}_{\omega},$…${\delta}_{\alpha},$…

that according to the theorem from p. 13 is countable and hence, necessarily stops at a definite $\alpha.$

This meant that the sequence of ordinals he had created must eventually terminate, else he would get an uncountable list of ${\delta}_n,$ which is a contradiction. Notice that this is not the same as saying that the list is finite, just that eventually the list of limit ordinals must end.

The following is the "theorem from p. 13" to which Schoenflies referred.

IV. Each infinite set $G$ of intervals of a continuous space $C_v,$ which lie exterior to each other or at most intersect at their borders, is countable.

Schoenflies didn’t give details of the proof, but the result is standard and can be left as an exercise to students. It often proceeds as follows: If one starts with a set of non-overlapping intervals, then each one must contain a rational number. By choosing a rational number $r_i$ in each $\delta,$ one can create a one-to-one correspondence between the intervals and a subset of the rational numbers. Because the rational numbers are countable, so is the subset, and hence so is the set of non-overlapping intervals.

In this case, it is not clear why this theorem is relevant. After all, the intervals ${\delta}_n$ can overlap! Schoenflies actually applied this theorem to the intervals $\left(a_i, a_{i+1}\right)$ writing, “Namely if $a_1$ > $a_2$$a_3$ . . . . is a sequence of positive numbers decreasing to zero without end, then the intervals whose content is between $a_{\nu}$ and $a_{\nu +1}$ …” [15, p. 13]. Because these intervals don’t overlap, they are countable and so are the $a_i.$ In Young’s proof in the next section we will see another trick that would have allowed Schoenflies to apply this theorem directly.

Let us return to the main theorem. After claiming that the sequence of ${\delta}_n$ is countable, Schoenflies argued that he could replace this infinite list of intervals with a finite list. He started by showing that he could cover $\left[a_{\omega}, a_1\right]$ by a finite number of intervals. Because $a_{\omega}$ is a limit point, then an infinite number of the $a_i$ lie within ${\delta}_{\omega}.$ In other words, there exists a number $\mu$ so that all $a_i$ with $i\ge\mu$ will lie within ${\delta}_{\omega}.$ Therefore $\left[a_{\omega}, a_1\right]$ will be covered by the finite set of intervals ${\delta}_1,$ ${\delta}_2,$ …, ${\delta}_{\mu},$ ${\delta}_{\omega}.$

Now this sequence of intervals can always be substituted by a finite set of analogous intervals. Returning first to point $a_{\omega},$ there is certainly a number $\mu$ so that all points $a_{\nu}, a_{\nu +1},$ ... lie within ${\delta}_{\omega}$ and hence, the interval $a_1$ ... $a_{\omega}$ is covered already by the intervals ${\delta}_1,$ ${\delta}_2,$…${\delta}_{\mu},$ ${\delta}_{\omega}.$

He argued that this process of reduction to a finite number of intervals does not work just when passing from the finite numbers to $\omega,$ but whenever $a_{\beta}$ is a limit ordinal - that is, a limit of a strictly decreasing sequence of left endpoints. The details of the induction are missing. Hallet filled in the holes, noting that, “he only proves the induction step from finite numbers to $\omega,$ and not in complete generality” [9, p. 22].

However, this is true for every point $a_{\beta}$ which is the limiting point of a sequence $a_{\alpha_1}, a_{\alpha_2}, a_{\alpha_3}, \dots, a_{\alpha_v},\dots$ so that the terminus of $\{a_{\alpha_v}\}$ satisfies that $a_{\alpha_{\omega}} = a_{\beta}.$ Assuming namely that every $a_{\alpha_{i+1}}$ from $a_{\alpha_i}$ on can be arrived at via a finite number of intervals, then this is also true of $a_{\beta}$ because again $a_{\beta}$ belongs to an interval ${\delta}_{\beta},$ and there is a definite $\mu$ with all points $a_{\alpha_{\mu}},$ $a_{\alpha_{\mu+1}},\dots$ belonging to $a_{\beta}.$ Now since the intervals $\delta$ should also contain the endpoints $a$ and $b,$ the claim is proved.

It is clear that this proof is very similar to Borel’s, though it does make a rather clumsy argument that the original cover need not be countable. Immediately upon completing the proof, Schoenflies stated that the theorem is also true in higher dimensions, and proceeded to sketch how the proof would proceed. It is different from the two dimensional argument of Cousin, for Schoenflies required the use of the one-dimensional theorem. Some of the details are missing, but it can make for a good project for students to unravel the proof.

It is not difficult to apply the same theorem to planes and spaces. Namely, if $a_1$ is now a point of a rectangle $H$ and ${\delta}_1$ the region around it, which for the sake of simplicity I shall consider as a square, then to every point $a_2$ on he perimeter of ${\delta}_1$ there also belongs such a square ${\delta}^{\prime}_1$ and it follows from the proven theorem, that there exists a finite number of squares such that all points $a_2$ on the perimeter of ${\delta}_1$ become interior points of one of these squares. There exists therefore in any case also a square ${\delta}_2$ which encloses ${\delta}_2$ such that all points within and on the perimeter of ${\delta}_2$ are covered by a finite number of squares. For this square there exists an analogous square ${\delta}_3$ etc. and the proof proceeds analogously to the proof above. Here as well every point of the perimeter of $H$ must belong to a region ${\delta}.$

As before, we now provide details of the argument that teachers may consider before presenting this argument in a classroom.

###### Background:
• If teaching this proof, one must cover completeness in the sense of the monotone convergence property or the Bolzano-Weierstrass property to show that the left endpoints have a limit.
###### Benefits:
• This proof closely follows that of Borel, though it does not require that the original cover be countable.
• Similarly to Borel’s proof, it offers a nice application of Cantor’s ordinal numbers.
• The outline of the generalization to two dimensions (or even more) would make a valuable project for students to work though.
###### Drawbacks:
• The step where Schoenflies proved that the covering must be countable is buried in a different section. And, it is not immediately clear how to apply that theorem. Details certainly need to be provided for a student to understand the full proof.
• As Hallet wrote, Schoenflies omitted several details in the induction step that an instructor would need to fill in.
###### Impressions:

As in the case of Borel’s original proof, this proof may be particularly relevant if the theorem is covered in a set theory or measure theory course.

Arthur Schoenflies (1853-1928) (Convergence Portrait Gallery)

###### Later Proofs:

A few years later in 1907, Schoenflies published another note, “Sur un théorème de Heine et un théorème de Borel” [16]. As we mentioned above, it was in this paper that he took the opportunity to defend his choice of attaching Heine’s name to the theorem. He also gave another proof, which is interesting as well. Around every point p of a closed set P, he defined ρ to be the greatest radius of an interval containing p. He then showed that the lower limit of all the ρ was not zero. He proceeded, saying “In fact, if this limit were zero, one could choose points p1, p2, …, pγ, in such a manner that the radii ρ1 > ρ2 > … > ργ … converge to zero. Let $p_{\omega}$ be a limit point of { pγ }; for this point there exists a radius ${\rho}_{\omega}>0$ and forcing the well-known contradiction.”

Schoenflies didn’t elaborate on the “well-known” contradiction, but it may proceed along these lines: let U be the open interval around $p_{\omega}.$ It has radius ${\rho}_{\omega}>0.$ Since $p_{\omega}$ is a limit point there are an infinite number of the points p1, p2, …, pγ in U. Those points form a subsequence of the original sequence, and so would have corresponding radii ρi converging to zero. However, because each point is in the interval around $p_{\omega},$ each ρi must be at least ${\rho}_{\omega},$ which is fixed so the radii ρi can’t converge to zero.

Once he knew that every point p is contained in an interval with positive radius > δ, Schoenflies could show a finite number would cover. This makes for a nice exercise for students. He also commented that, “This demonstration is exactly the same, whether the supposed set of domains is countable or not” [16, p. 23].

Nicole R. Andre (Wittenberg University), Susannah M. Engdahl (Wittenberg University), and Adam E. Parker (Wittenberg University), "An Analysis of the First Proofs of the Heine-Borel Theorem - Schoenflies' Proof," Convergence (August 2013), DOI:10.4169/loci003890