Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

On 10/24/24 7:07 AM, WM wrote:

On 24.10.2024 05:04, Richard Damon wrote:

On 10/23/24 10:46 AM, WM wrote:

On 23.10.2024 13:37, Richard Damon wrote:

On 10/22/24 12:12 PM, WM wrote:

On 22.10.2024 18:03, Jim Burns wrote:

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∀n ∈ ℕ:  2×n ∈ ℕ

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Not if all elements are existing before multiplication already.

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IF not, then your actual infinity wasn't actually infinite

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It is infinite like the fractions between 0 and 1. When doubling we get even-numerator fractions, some of which greater the 1.

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But from 0 to 1 isn't an infinite distance

 Measured in rational points it is infinite.

But values are not measured in "rational points".
So, your arguement doesn't make sense.

 

At 1/2, we double to hit the boundry.

 There we have crossed many dark fractions already.
Same at ω/2.

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But, with the Natual Numbers, there isn't a value that is 1/2 of the "highest value" since there isn't a highest number that is a Natural Number, since every Natural Number has a successor that is higher.
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It is claimed that there are all numbers. "That we have for instance when we consider the entirety of the numbers 1, 2, 3, 4, ... itself as a completed unit, or the points of a line as an entirety of things which is completely available. That sort of infinity is named actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p. 167]

So?
If you HAVE all the numberes, 1, 2, 3, 4, ... that set goes on FOREVER and doesn't have a upper end.
If you have the COMPLETE unit, it doesn't have a highest number.
Your operation of doubling the values on the line from 0 to 1 isn't operating the property that that set is infinite on, so doesn't follow the law of the infinite.

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Your "actual infinity" seems to be just an unimaginably large value, not infinite, as your actual infinity has an end, it has an element without a successor, so it isn't the set it claims to be.

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The completed infinite cannot avoid to be complete. But it is infinite because the end cannot be determined because of the dark domain.

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No, the end cannot be determined, because it isn't there.
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There is nothing about being complete that means it needs to have an "end"

 Whatever, it is complete and all its numbers can be doubled. Some are resulting in larger numbers tha have been doubled.

Nope, as every number (A Natural Number) doubles to another number in that set (The Natural Numbers) so you never left the set.
It may seem agaist your intuition, but that is because you are only thinking in terms of finite sets.

Regards, WM