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Replies: 62   Last Post: Feb 24, 2014 10:17 PM

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 thenewcalculus@gmail.com Posts: 1,361 Registered: 11/1/13
Posted: Feb 23, 2014 7:20 PM

On Sunday, 23 February 2014 21:56:17 UTC+2, Julio Di Egidio wrote:

> That is not about proof, cardinality in terms of bijections is rather a
> definition.

A very bad definition! It is ill formed.

> That said, yes, I think it is sensible to say that the
> collection of even natural numbers is equinumerous to that of all natural
> numbers, but I have already characterised in which sense I mean that: in the
> sense that both are *endless*, nothing less and nothing more.

It's completely delusional to say that the collection of even natural numbers is equinumerous to that of all natural numbers. Why so? Let's see of I can make this very simple.

Firstly, I want to discuss a bijection between two subsets of the so-called "real numbers". Let's take the real intervals (0,1) and (0,5). Since a bijection exists, these sets have the same cardinality according to the definition. But what is being mapped? Numbers or points? In actual fact, points are being mapped, not numbers. In between any two points there are infinitely many other points. If indeed (0,1) and (0,5) have the same cardinality, then how can it be that the length of (0,5) is greater than (0,1)? When you think of this in terms of the fallacial density argument, remember that we are talking about the points between 0 and 1, and between 0 and 5. You can say NOTHING about the number of points between any two points. Sure you can create a mapping function, but that does not reify the points.

Secondly, the natural numbers CAN be reified as points on the number line. What this means is that we can confidently state there are no natural numbers between two consecutive natural numbers. Now think again of what I told you in the previous comment:

1 2 3 4 5 .... is the set of natural numbers. Remove the even points and you will have less points left, bijection or no bijection. In fact you can use your invalid n and n+1 argument here if you wished. You could say that for any n consecutive elements, if we remove the even natural numbers from these, we'll have [n/2] natural numbers left. See where I am leading?

> Again, just as much as there is an infinity of natural numbers: i.e. it is
> another if-then.

So let me ask you this: In evaluating a conditional such as if-then, do you conclude the "then" part will be true if the "if" part is false? Of course not. In software development we have lots of if-then conditionals. The "then" part is only executed on the premise that the "if" part is true. In your example, the "if" part is clearly FALSE.