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# Fallacy by deflection

A well-known professor claimed that there was no possibility that the set A as in the Courant-Robbins contradiction and Lebesgue Measure could be the entire unit interval 0 to 1. (Footnote: See the appendix below Defining a set over the interval 0 to 1 for a brief synopsis.) This is the argument he used:

We define a list (Footnote: See One-to-one correspondences.) of rational numbers by taking a pre-existing list of rationals and we define that if there is a rational in the list that is close to one given specific irrational number, then that rational is pushed further down the list. The closer that rational is to that irrational number, the further it is pushed down the list. More formally the definition is:

Take any arbitrary irrational r where 0 < r < 1. Given that a1, a2, a3, … is an enumeration of the rationals in [0,1] (the interval 0 to 1, including 0 and 1), we create a new enumeration, b1, b2, b3, … where for all n, bn is the first ai whose distance from the chosen r is greater than 110n and is different from all bm for m < n (i.e., different from all bm already in the list).

While this definition is fairly informal, there is no difficulty in making it more formal, viz:

Given any arbitrary irrational r where 0 < r < 1, and any arbitrary enumeration a1, a2, a3, … of the reals over [0,1], a new enumeration b1, b2, b3, … is defined by:

n, ∃i (bn = ai )

if and only if

m, m < n, bm ≠ ai ∧ ¬[ai110n  < r < ai + 110n ]

The professor then continued:

When we combine the definition of this list with the definition of decreasing intervals 110, 1100, 11000, … 110n, … we have a new definition.

This defines a set A*. It is impossible for r to be in the set A*. Therefore r is not in the set A* but it is an single point, therefore there cannot be an interval to the right or left of that point that is an open interval with a rational as its endpoint, since the endpoint would have to be r, and r is irrational.

## Infinite sets of single points

Well, there is an obvious fallacy in the above - you cannot prove a generalization from a finite number of instances, never mind a single instance.

But besides that, the argument throws up more problems than it solves. While it shows that, while you can define a list of rationals so that it will exclude given irrational numbers, it also shows that every such excluded irrational is associated with an infinite sequence of rational numbers. Furthermore, if there is more than one such excluded irrational, it is clear that each such excluded irrational:

1. is an single point and
2. is associated with its own unique infinite sequence of rational numbers.

And that means that if there could be infinitely many such excluded irrational numbers, then that infinity could not in any sense be a “bigger” infinity than that of the rational numbers.

But, according to conventional Lebesgue measure theory, the set of all these single points not in the set A* is a set that has a measure of at least 89, but any infinite set of single points that has a finite definition (Footnote: e.g., the rational numbers, the square roots of prime numbers) has a measure of zero.

Note that the professor’s definition tells us nothing more about membership of his new set A* other than it is the original set A less the single irrational r - it tells us nothing about other members of the set A*.

## Limiting Conditions

Clearly, any such point r as above cannot be in an non-degenerate interval that is not in the set A* since any such interval would also contain rationals, which by definition, must be in the set A*. Hence any such point must be a degenerate interval which is a closed interval whose endpoints are that point itself.

Now, the definition of the set A* is that all intervals of A* are open intervals with rationals as the endpoints of each such interval. Any interval that remains after some interval of  A* is removed from the interval 0 to 1 must be a closed interval that has rational endpoints that are not in A*.

So how can the situation arise where we are left with a set of at least one (degenerate) interval that is not in  A*, but it is a closed interval with irrational endpoints?

Clearly, there is no natural number for which there is a corresponding interval of  A* that is removed from the interval 0 to 1, and which leaves behind a closed interval with irrational endpoints.

All that this is telling us is that the Platonist notion that the totality of infinitely many all real numbers between 0 and 1 actually exist simultaneously in some non-physical sense is untenable. On the other hand, we can acknowledge that there is no termination of the selection of intervals of the set A*, so that a contradiction can never arise. But we then also have to acknowledge that the definition does not define a definitive set membership. That means that we need to define a limiting condition to define a certain set of points, where the selected number r cannot belong to the that set, and where the defined endpoints of removed intervals do not hold for that limiting condition.

Footnotes:

#### Defining a set over the interval 0 to 1

First, definitions of an open and a closed interval:

An open interval is an interval that does not include the endpoints that define that interval (for example the open interval whose endpoints are 13 and 12 is the set of all points between 13 and 12 but not including the points 13 and 12).

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

Now, define a set A in terms of ever decreasing intervals that are associated with the list:

We start with the closed interval between 0 and 1. Now take a suitable list of the rational numbers between 0 and 1 (see below). Then, going through this list of rational numbers, for the first rational we define an open interval 110 wide with that rational at the midpoint of the interval; our set now includes all the numbers in that interval (not including the endpoints). For the next number, define an open interval 1100 wide with that rational at the midpoint of the interval; we add those numbers to our set. For the next number, define an open interval 11000 wide with that rational at the midpoint of the interval; we add those numbers to our set. And so on, with each subsequent open interval being 110 of the length of the previous interval.

Given this definition, either the set A is the entire interval between 0 and 1, or else there are irrational single points that remain in the interval 0 to 1. Either way, since single points do not have any length, the set A has a total measure of 1. But Lebesgue measure theory claims that the measure of the set A must be less than 19. For more on this, see Lebesgue Measure Theory. But some people do not like this result and try to find arguments against it, such as the one described on this page.

#### A specific listing of rational numbers

Some people have suggested that they can circumvent the contradiction by using a list (see also One-to-one correspondences and Listing the rationals) of the rationals that are defined in terms of various conditional requirements, which render the enumeration and the sequence of intervals interdependent. Rather than trying to construct a set of rules as to which enumerations are applicable, all that is required is one specific enumeration. We can define that the set A is to be given by one specific enumeration using the pattern of rationals:

 1⁄2 1⁄3 1⁄4 1⁄5 1⁄6 … 2⁄3 2⁄4 2⁄5 2⁄6 … 3⁄4 3⁄5 3⁄6 … 4⁄5 4⁄6 … 5⁄6 …

This gives a list that begins as 12, 13, 23, 14, 34, 24, 15, 25, 35, 45,16, 26, 36, 46, 56,… Note that this gives duplicates like 24, 321, but the important fact is that the enumeration includes every rational between 0 and 1 (not including 0 or 1). Clearly, there can be infinitely many similar enumerations based on similar definitions. Note that this enumeration follows a pattern that for each subsequent denominator, the values run from the lowest to the highest value of the numerator.

The enumeration can be represented by an algorithm as follows, where the calculation of the nth rational does not require the calculation of any of the rationals prior to n in the enumeration:

1. Let t = 0
2. If 1 + 8(n + t) is a Natural number, then:
3. m = (−1 + √1 + 8(n + t) )/2 and the nth rational is (mt)/(m + 1)
4. Otherwise let t = t + 1 and repeat from step 2.

This can easily be made into a program as has been done here:

Enter the desired number in the enumeration of the rationals (between 1 and 1000000):

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## NEWS

### Paper on the diagonal proof

There is now a paper that deals with the matter of language and the diagonal proof, see On Considerations of Language in the Diagonal Proof.

### Other recently added pages

The Myths of Platonism

The Platonist Rod paradox

The Balls in the Urn Paradox

### Lebesgue Measure

There is now a new page on a contradiction in Lebesgue measure theory.

### Easy Footnotes

I found that making, adding or deleting footnotes in the traditional manner proved to be a major pain. So I developed a different system for footnotes which makes inserting or changing footnotes a doddle. You can check it out at Easy Footnotes for Web Pages (Accessibility friendly).

### O’Connor’s “computer checked” proof

I have now added a new section to my paper on Russell O’Connor’s claim of a computer verified incompleteness proof. This shows that the flaw in the proof arises from a reliance on definitions that include unacceptable assumptions - assumptions that are not actually checked by the computer code. See also the new page Representability.

### Previous Blog Posts

For convenience, there are now two pages on this site with links to various material relating to Gödel and the Incompleteness Theorem

– a page with general links:

– and a page relating specifically to the Gödel mind-machine debate:

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