Understanding sets of decreasing intervals

Page last updated 02 Sep 2021

Some people appear to have difficulty with understanding a logical analysis of cases of decreasing intervals such as the Courant and Robbins contradiction and Lebesgue measure. I have explained in Understanding Limits and Infinity how the application of limits provides a logical result without any concomitant contradictions, but this does not appear to satisfy some readers. So I have written this page to show why the contradictions that are inherent in conventional treatments of such cases cannot be resolved without attending to how infinity is dealt with in the definitions.

We will deal with the definition of a set A in terms of ever decreasing intervals; the definition is similar to that on the page Lebesgue measure, except that we do not specify whether the intervals are open or closed:

We note here some definitions:

Open interval: An open interval is an interval that does not include the endpoints that define that interval. (Footnote: An example of an open interval is one whose endpoints are ¹⁄₃ and ¹⁄₂ is the set of all points between ¹⁄₃ and ¹⁄₂ but not including the points ¹⁄₃ and ¹⁄₂.)

Closed interval: A closed interval is an interval whose endpoints are included in the interval.

Complete Interval: An interval of A is a complete interval if it is not a sub-interval of any interval of A apart from itself. (Footnote: For example, if a set X consists of two overlapping closed intervals, the interval with endpoints ¹⁄₅ and ¹⁄₃ , and the interval with endpoints ¹⁄₄ and ¹⁄₂ , then X consists of the complete closed interval with endpoints ¹⁄₅ and ¹⁄₂ .)

Degenerate interval: A degenerate interval is an interval that consists only of a single point

At this point we note that there cannot be any degenerate open interval; suppose for a moment that there could be such an interval - then its endpoints must either be different or be identical. If its endpoints could be different, then ipso facto, they would define a non-degenerate interval which obviously could not be a degenerate single point. But on the other hand, if the endpoints could be identical, then by the definition of open, the endpoints cannot be included in the interval - and that means that there are no points that could be in the supposed interval, and so there cannot be any such interval.

Endpoints

We can also observe that, ignoring for the moment the question of which points belong to which set, every point is:

1. the left endpoint of infinitely many closed intervals, and
2. the left endpoint of infinitely many open intervals, and
3. the right endpoint of infinitely many closed intervals, and
4. the right endpoint of infinitely many open intervals.

So, for example, if a point is the right endpoint of a complete open interval of a set X, that point is the left endpoint of infinitely many closed intervals. And since, by definition, that point is not a point of the set X, then it can only be a point of some other set Y, and so it must be the left endpoint of some closed interval of the set Y - the other points of that interval are given by the definition of Y - it could be a single point degenerate interval or it could be a non-degenerate interval.

The intervals of B

Consider what can be in the set B. Any set of intervals has to consist of either degenerate intervals, non-degenerate intervals, or some combination of both. But there cannot be any non-degenerate intervals in the set B, since every non-degenerate interval must include rationals, which must be in the set A. (Footnote: Given the two endpoints of a non-degenerate interval, then since they are different, there are infinitely many points in that interval, and which must include rationals. ) Hence the only possibility is that if there are points in the set B, then they must all be degenerate single point closed intervals. Furthermore they must all be irrational, since all rational points must be in the set A.

Endpoints of complete intervals of A

Looking at the definition of the set A, we can see that there cannot be any complete intervals of A that are degenerate single points, since every interval of A must have a midpoint that is rational which has an associated number n, and it must have a well-defined non-zero measure which is ¹⁄_10ⁿ.

Since every rational is the midpoint of an associated interval ¹⁄_10ⁿ then the endpoints of every interval of A must also be rational (a rational plus a rational is always rational). A complete interval of A is composed of sub-intervals that follow the definition of intervals of the set A and which overlap. Since every sub-interval has rational endpoints, then every complete interval of A must also have rational endpoints.

At this point we may observe that we have not specified in the definition of the set A whether the intervals are open or closed; we will look at both options. This means that the endpoints can either be included in that interval or not included in that interval. But we also know that for either option, the endpoints of a complete interval of A cannot be rational. We know this, since if an endpoint of a complete interval could be rational, it would be a midpoint of some non-degenerate interval of A whose measure is ¹⁄_10ⁿ - which means that it overlaps the supposed rational endpoint of the supposed complete interval. Hence the endpoints of any complete interval of A cannot be rational.

A contradiction

But now have a contradiction; the endpoints of any complete intervals of A must at the same time be rational and not rational. (Footnote: All intervals are either closed degenerate, closed non-degenerate or open non-degenerate. Hence all complete intervals have definitive endpoints, regardless of whether they are closed or open, and such points must be either rational or irrational.) We will consider later the reasons for the contradiction. But before doing so, we will set this contradiction aside for the moment and analyze the cases where we assume that the endpoints of the complete intervals of A can be irrational.

At this point we recall that we did the above analysis under the assumption that we could ignore the contradiction that the endpoints of the intervals of A are defined as being rational, while the definition also results in the conclusion that endpoints of any complete interval of A cannot be rational.

This is a contradiction that demonstrates that there can be no set A that satisfies the above definition of a set A, regardless of whether the definition is in terms of open or closed intervals. Furthermore, it cannot be an ‘empty set’, (Footnote: For more on the ‘empty set’, see the page: The ‘Empty Set’.) since the interval for the first n in the enumeration is in the set, and the other intervals are defined as being added to that set, and points are never removed.

It is easy to see that by adding to the definition some finite upper limit on the value of n, we can have a definition that defines a perfectly well-defined set. So it is clear that the cause of the contradiction lies with the failure to define what happens with limitlessly many n, as for the definition on the page Lebesgue measure. I explain in Understanding Limits and Infinity how the inclusion of limits in the definitions of the set A, either with open intervals or closed intervals, gives logically coherent definitions without any resultant contradictions.

The Origins of Transfinite Numbers shows how misunderstandings regarding the unions or intersections involving infinitely many sets originated and later became ingrained into mathematics.