New Anti-Cantor Attack
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From: Jim Burns on 1 Jun 2010 13:57 |-|ercules wrote: > "Transfer Principle" <lwalke3(a)lausd.net> wrote > >> And even though Herc/Cooper turned out to be >> not worth defending in the end, >> that still doesn't mean that I'm going > [...] > > I'm going to substitute me not being worth > defending with my case not being worth defending, > and leave you be with this. For once, I heartily agree with you, Herc. (With a "Yeah! Verily, yeah!"). The point (my point, Herc's point) is that the worth of the argument is not determined by the worth of the arguer. (Savor the moment, Herc. It may not come again for a long while.) Jim Burns
From: Bart Goddard on 1 Jun 2010 14:18 Jim Burns <burns.87(a)osu.edu> wrote in news:hu3hll$jqd$1 @charm.magnus.acs.ohio-state.edu: > The point (my point, Herc's point) is that > the worth of the argument is not determined by > the worth of the arguer. > How do you propose that a worthless arguer come up with a good argument? Maybe "determined" is too strong a word, but at the very least, the correlation coefficient here is .999.... B. -- Cheerfully resisting change since 1959.
From: George Greene on 1 Jun 2010 14:56 On May 29, 10:58 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Here's my equivalent proof of uncountable infinity. It's NOT "equivalent", but it does have a very similar structure. It does similarly prove that something newer and bigger exists. The only problem is, what your proof proves the existence of IS NOT a natural number. Your proof is a proof that EVERY natural number FAILS to be something (i.e., the biggest). > > Let's assume an enumeration of naturals exists, call it N. > > N= { > 1, > 2, > 3, > 4, > ... > > } > > Let's calculate a new natural MAX+1. That is not possible, because YOU HAVE NOT SAID what "MAX" is. Is MAX supposed to be the maximum natural number? The problem with that is, MAX *did not show up* on your "enumeration" of the naturals above. Your enumeration ended with "...", and nobody really knows what you meant by that. If you meant the usual thing, then, obviously, that sequence DOES NOT HAVE a last or maximum element, so your calculation is impossible. If you wanted to ACTUALLY BE ABLE to start this "calculation", then your enumeration WOULD have had to look like > N= { > 1, > 2, > 3, > 4, > ... > MAX > } If you are going instead to say > That is 4+1 = 5 > in this finite subset example. Then your enumeration has to look like > N= { > 1, > 2, > 3, > 4 = MAX > } WITHOUT the "..." . But see, your problem is, you're stupid, so you couldn't even state YOUR OWN example accurately. > Voila 5 is a new number not in N > > Therefore no matter how big N is there is always a new element > that can be listed Right. Up to this point, you are telling the truth as best you know how. If you just knew WHERE TO GO from here, then everything would be OK. You have been constructing a proof that is going to use the inference rule called universal generalization. It is going to say something about EACH AND EVERY natural number. What you are (correctly) saying it says is that, for every natural number, there is a biggER one, and therefore, for every natural number n, n is NOT the biggest one (n =/= MAX ; in point of fact, MAX(N) simply does not exist at all). > and therefore the size of the set N is bigger than infinity. Now, you're just being an idiot. Nothing in your proof said ANYTHING WHATSOEVER about SET SIZES! YOU WERE ONLY talking about numbers and enumerations! You have (sort of) correctly proved that NO enumeration naturals WITH a MAX element is an enumeration of ALL of them, but that simply does NOT SAY anything about "bigger than infinity"! If you WANTED to say something about set SIZE then you wuold NEED to FIRST actually SAY something about set SIZE, namely, "The SIZE of the set of the first N naturals (starting with 1 and ending at n) IS n." Had you said THAT, THEN you could say, "For no n is n the size of the set of ALL the naturals". This REALLY WOULD PROVE that there was a "new" size that was "bigger" than any NATURAL size. But that would NOT be proving that anything was "bigger than infinity" -- that would just be proving that the size of N *is* infinity. And it really would be a whole new kind of [cardinal] "number".
From: George Greene on 1 Jun 2010 14:59 On May 29, 10:58 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Cantor's second proof of 'uncountable infinity' is based on trying to enumerate the powerset of naturals. > > e.g. > P(N) = { > 1 - {1}, > 2 - {1,2}, > 3 - {2}, > 4 - {1,2,3} > ... > > } > > The set of indexes that aren't members of their subset is > {3,4} > in this finite subset example. > > And voila, {3,4} is a new subset not present in P(N). > > Therefore no matter how big P(N) is there is always a new element > that can be listed and therefore the size of the set P(N) is bigger than infinity. Your use of "therefore" here is incorrect. This IS NOT HOW our version of the proof goes. What happens after "there is always a new element of p(N)" is, THEREFORE, THE OLD list WAS NOT COMPLETE. We make a generalization about the LISTS. What this proof proves is that NO list is EVER a complete list of all the subsets. But these ARE INFINITE lists! NO countably infinite list is EVER long enough (they are ALL too short), therefore, the set being listed IS BIGGER than countABLE. This is the EXACT same thing that is going on in your proof: No FINITE list is ever long enough (to list all the naturals), therefore the set of naturals IS BIGGER than any finite number (i.e. it is infinite). BOTH proofs create a new bigger number in a similar way.
From: George Greene on 1 Jun 2010 15:02
On May 29, 11:36 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Yes I know it's such a simple yet equivalent logical deduction of 'bigger than infinity' > and really shows how dumb Cantor subscribers are to miss that, No, it isn't. Your second proof is a deduction of "bigger than FINITE" because all the lists you are generalizing over ARE FINITE -- they have a last MAX element. Even though you were too stupid to draw it that way. If the list actually is a list of all the naturals in order then IT HAS NO MAX, and your proof can't even get started. > but the fundamental mathematical truths are generally quite simple. Of course they are, yet despite this, you are too stupid to apply them. |