Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary, effectively)

On 12/31/24 1:05 PM, WM wrote:

On 31.12.2024 16:37, Richard Damon wrote:

On 12/30/24 4:38 PM, WM wrote:

On 30.12.2024 17:08, Richard Damon wrote:

On 12/30/24 3:44 AM, WM wrote:

On 30.12.2024 01:36, Richard Damon wrote:

On 12/29/24 5:31 PM, WM wrote:

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My theorem: Every union of FISONs which stay below a certain threshold stays below that threshold.
Find a counterexample. Don't claim it but prove it. Fail.

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But what does that prove?

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That proves the existence of dark numbers.

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How?
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It shows that there are numbers you didn't look at, but you admit to not looking at all the numbers.

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Neither can you look at more numbers.

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Of course I can look at "more" numbers, as I can look at the one after where you stopped.

 I don't stop but know that every FISON is finite as its name says and covers at most 1 % of ℕ. If you don't agree find a counterexample.
 

But then you think that a finite combination of these finite sets could possible by an infinite set, or even the results of combining *ALL* of the members of the infinite set of FISONs. Even though each one is finite, there is an infinite number of them, so words like "Every" or "All" require a logic that can actually handle that infinity, which yours doesn't, as you can deal with all the members of an infinite set individually with a finite process.

If ℕ is an actually infinite set then it is not made by FISONs.

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That is what I have been telling you for a long time.

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But there is no Natural Number that isn't in a FISON, which is NOT what you have been saying.

 You cannot find a natural number that isn't in a FISON. But there must be more because FISONs cover less than 1 % of ℕ.

Right, but there are an INFINTE number of FISONs, just like there are an INFINITE number of natural numbers, so any system that can only handle a finite number of operations can't be expected to fully handle the set of FISONs or Natural Numbers.

 

The problem is your logic can't have "Every" FISON, because "every FISON" is an infinite set of them,

 Every mathematician knows that every FISON is finite as its name says and covers at most 1 % of ℕ.

But, when you actually take into accout EVERY FISON, the full INFINITE set, you get to them all. Of course, your problem is that you logic system can't actually handle EVERY FISON, just ANY FISON, and thus can't determine the property of the union of EVERY FISON.
You just don't understand the nature of the universal qualifier as applied to truely infinite sets.

 Regardes, WM