Re: Incompleteness of Cantor's enumeration of the rational numbers

On 11/24/24 10:23 AM, WM wrote:

On 24.11.2024 15:40, joes wrote:

Am Sun, 24 Nov 2024 11:31:00 +0100 schrieb WM:

On 24.11.2024 03:22, Richard Damon wrote:

On 11/23/24 4:11 PM, WM wrote:

>

Cover the unit intervals of prime numbers by red hats. Then shift
the red hats so that all unit intervals of the positive real axis
get red hats.

And you can, as the red hat on the number 2, can be moved to the
number 1 the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3

>

It fails in every step to cover the interval (0, n] with hats taken from
this interval.

Nobody cares about a nonbijection from N to N. We are interested in a
bijection from N to P, or {0, 1, ..., n} to {p0, p1, ..., p_n}

 That can be excluded because there are too few hats in all intervals.

But the bijection is for INFINITE sets, not finite sets, so you are just showing you believe in strawmen.

>

Yes, for every n that belongs to a tiny initial segment.

No, for EVERY n. Show one that it doesn't work for!

The complete covering fails in every interval (0, n] with hats taken
from this interval.

Think about it the other way around: when we take the primes and
number them, we are never done, because there are inf.many primes.
Therefore we need the set of all naturals, N, to count them.

 Even the naturals divisible by 10000000 would suffice if Cantor was right. But he is not.

But he is, and you are not because you mind has been exploded by your use of incorrect logic.

 

so all the numbers get covered.

No.

WHich one doesn't.

Almost all. For every interval (0, n]
the relative covering is 1/10, independent of how the hats are shifted.

As above, nobody is arguing that the primes are somehow more infinite.

 They are less than the naturals.

Nope, They form a 1:1 so are of the same size, Aleph_0, you just don't understand the mathematics of such a thing, causing your mind to be exploded.

>

This cannot be remedied in the infinite limit because outside of all
finite intervals (0, n] there are no further hats available.

Yes it can, if you start out with a proper finite bijection.

 If it could exist, then all hats could cover all naturals. But that is excluded by the relative covering of 1/10 in all intervals (0, n].

Nope, that is just your mind being exploded by your use of invalid logic on an infinte set.

>

On the other hand, we cannot find a first n that cannot be covered by a
hat. This dilemma cannot be resolved by negating one of the two facts.
It can only be solved by dark numbers.

Or getting over your wrong intuition.

 The limit of a constant sequence is its constant. No intuition.

Only if the results applies to the infinite.
Sorry, your logic is just bad, being based on an incorrect "intuition"

 Regards, WM

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