Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.logicDate : 29. Nov 2024, 23:05:25
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <abd1031a1be7fec415c1e8b43216d46c31493427@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
User-Agent : Mozilla Thunderbird
On 11/29/24 4:39 PM, WM wrote:
On 29.11.2024 21:07, Richard Damon wrote:
On 11/29/24 2:28 PM, WM wrote:
Analysis is no a lie.
>
Bad Analysis, like what you do is.
The limit of the infinite sequence 1/9, 1/9, 1/9, ... is 1/9. Nothing is clearer than that.
Which doesn't actually mean anything.
The limit x-> 0 of 0^x is 0 (as it is for all x)
The limit x->0 of x^0 is 1 (as it is for all x)
But 0^0 isn't defined, even though both of the limits seem to apprach it.
You can only use the limit as the final result if it actually applies.
Since the Infinite set is actually the same kind as any of the finite sets in the sequence, the limit doesn't apply
No, I am using a very simple and sound rule. If all hats of finite intervals (0, n] fail to cover more than 1/10, then it is impossible to cover more than 1/10 of the whole set ℕ because beyond all finite intervals and all finite n, there is no supply of black hats.
>
So, you admit to MAKING UP your rules based on your own ideas
This chain of arguing is irrefutable by consistent thinking.
No, it is based on inconsistent thinking that has blown up your logic to smithereens causing you to not know what it true.
You are just working off naive mathematics that was proving incorrect, and are stupidly sticking to it.
Sorry, but you are just proving your stupidity.
Regards, WM