Criticism of the doctrine of infinity

[Kekkuli]

On the one hand, Cantor showed that not all real numbers can be enumerated, while on the other hand he showed that rational numbers can. Cantor demonstrated this with a grid. In the picture below, a natural number (yellow) is assigned to each rational number in order, but since the natural number is also a rational number, a fractional number that has not yet been listed has been constructed in each listing step. In the picture below, the largest number among the rational numbers is 11/1=11, while the largest natural number is already 45, it follows that all fractions can never be listed. Enumeration always produces only more numbers that have not yet been enumerated.

 

[Mark44]

On the one hand, Cantor showed that not all real numbers can be enumerated, while on the other hand he showed that rational numbers can. Cantor demonstrated this with a grid. In the picture below, a natural number (yellow) is assigned to each rational number in order, but since the natural number is also a rational number, a fractional number that has not yet been listed has been constructed in each listing step. In the picture below, the largest number among the rational numbers is 11/1=11, while the largest natural number is already 45, it follows that all fractions can never be listed. Enumeration always produces only more numbers that have not yet been enumerated.
View attachment 338495

Your criticism is incorrect. Consider the following grid. Only part of it is shown, as the rows extend infinitely far off to the right and there are an infinite number of rows.
$\begin{vmatrix}1/1 & 1/2 & 1/3 & 1/4 & 1/5 & \dots \\ 2/1 & 2/2 & 2/3 & 2/4 & 2/5 & \dots\\ 3/1 & 3/2 & 3/3 & 3/4 & 3/5 & \dots \\ \cdot & \cdot & \cdot & \cdot & \cdot & \dots\\ \cdot & \cdot & \cdot & \cdot & \cdot & \dots\\ \cdot & \cdot & \cdot & \cdot & \cdot & \dots\\ \end{vmatrix}$

Surely you would agree that this would list all the positive rational numbers. To pair them with the positive integers, weave through them just as you showed in your diagram. If you want to count 0 and the negative rationals, create a similar grid with the negative rationals. You could then start the count with 0 as the first in your list, and then hop back and forth between the two grids, thereby placing all of the rational numbers in one-to-one correspondence with the counting numbers.

 

[Kekkuli]

Your criticism is incorrect. Consider the following grid. Only part of it is shown, as the rows extend infinitely far off to the right and there are an infinite number of rows.
$\begin{vmatrix}1/1 & 1/2 & 1/3 & 1/4 & 1/5 & \dots \\ 2/1 & 2/2 & 2/3 & 2/4 & 2/5 & \dots\\ 3/1 & 3/2 & 3/3 & 3/4 & 3/5 & \dots \\ \cdot & \cdot & \cdot & \cdot & \cdot & \dots\\ \cdot & \cdot & \cdot & \cdot & \cdot & \dots\\ \cdot & \cdot & \cdot & \cdot & \cdot & \dots\\ \end{vmatrix}$

The fact that the set of rational numbers would be numberable would mean that there would be a bijection from the set of rational numbers to the set of natural numbers. But you have, for example, the numbers 1/1, 2/2, 3/3,...,n/n=1 among the rational numbers, but the bijection should have a goal and the starting set should have a number only once. Your grid is wrong.

 

[FactChecker]

The fact that the set of rational numbers would be numberable would mean that there would be a bijection from the set of rational numbers to the set of natural numbers. But you have, for example, the numbers 1/1, 2/2, 3/3,...,n/n=1 among the rational numbers, but the bijection should have a goal

What do you mean that it "should have a goal". Normally, bijections have no motivation or emotions at all. ;-)

and the starting set should have a number only once. Your grid is wrong.

So if you just skip any later duplicates, doesn't that leave you with a bijection between the rational numbers and the natural numbers?

 

[Kekkuli]

Well, if there was a bijection f from the set of natural numbers to the set of rational numbers, would you tell me what f(4895569879832983429893449999) is? You don't know it, and you can't find it out without an algorithm, and algorithms are always finite. There is no clause.

 

[Mark44]

Well, if there was a bijection f from the set of natural numbers to the set of rational numbers, would you tell me what f(4895569879832983429893449999) is?

I can't tell you off the top of my head which rational number pairs with the integer you apparently typed at random, but because of the grid pattern, it wouldn't be hard to find a formula that gives it.

You don't know it, and you can't find it out without an algorithm, and algorithms are always finite. There is no clause.

What does having an algorithm (or not) have to do with anything? And so what if algorithms are finite? This doesn't make any sense.

 

[FactChecker]

Well, if there was a bijection f from the set of natural numbers to the set of rational numbers, would you tell me what f(4895569879832983429893449999) is? You don't know it, and you can't find it out without an algorithm, and algorithms are always finite. There is no clause.

Nobody said that the bijection has to be a simple one that is easy to follow. But it is easy to see that there is a bijection. That is good enough. QED

 

[Mark44]

And with that last comment, the thread is closed.