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From: albstorz on 27 Oct 2005 11:02 Randy Poe wrote: > albst...(a)gmx.de wrote: > > My argumentation is very easy: > > Every nat. number represents a set. If you look at the first 100 nat. > > numbers, the 100th nat. number "100" represents the set {1, ... , 100}. > > As this holds for every nat. number, if there are infinite nat. numbers > > there must be a infiniteth nat. number representing this set. > > No, there musn't. There is no logical basis on which to draw > this conclusion from your premise. Yes, it must. You can derive from the peano axioms or settle an equal definition: nat. numbers are this, what count elements of sets. If there is a congregation which elements you can't count you have either no size or no set. Regards AS > > > But the definition of the nat. numbers with complete induction leads to > > the consequence, that there could not be an infinite nat. number. > > Correct. Your first "conclusion" is merely an assertion, not > based on axioms. The only rule you're using is "there must", > the same thing Tony Orlow does: shout loudly when you don't > have a mathematical basis for "must". > > There "must" based on WHAT? WHY must there? > > - Randy
From: Randy Poe on 27 Oct 2005 11:05 albst...(a)gmx.de wrote: > Randy Poe wrote: > > albst...(a)gmx.de wrote: > > > My argumentation is very easy: > > > Every nat. number represents a set. If you look at the first 100 nat. > > > numbers, the 100th nat. number "100" represents the set {1, ... , 100}. > > > As this holds for every nat. number, if there are infinite nat. numbers > > > there must be a infiniteth nat. number representing this set. > > > > No, there musn't. There is no logical basis on which to draw > > this conclusion from your premise. > > Yes, it must. You can derive from the peano axioms or settle an equal > definition: nat. numbers are this, what count elements of sets. > If there is a congregation which elements you can't count you have > either no size or no set. Which Peano axiom do you think justifies "If there are infinitely many natural numbers, there must be an infinitieth natural number representing this set." Again, you resort to "there must" and "you can derive" without showing an actual process of deduction. There isn't one. This conclusion does not follow. Non sequitur. Literally. - Randy
From: Tony Orlow on 27 Oct 2005 11:07 Randy Poe said: > > Tony Orlow wrote: > > Virgil said: > > > To declare something does not make it so, at least outside of TOmatics. > > It's my bijection, after all. I can declare it to be whatever I wish. > > You can't declare it to be a bijection unless you prove it > to be so. > > > Go ahead > > and rain on my parade, but don't move the police barriers. > > > > > > Outside of TOmatics, declarations require proofs before they need be > > > accepted. > > Not when you are defining a bijection. > > Yes, when you are defining a bijection. > > > What do you prove when you say the evens are mapped with f(x)=2x? > > That this map is bijective. It isn't assumed. It has to be > proved. All of us who offer it as a bijection are fully > prepared to back up that claim with a proof. We don't > claim it is "obvious" or "needs no proof" or "is a bijection > because we put it on the table with a label THIS IS A > BIJECTION." It's a map. If you aren't sure it's bijective, > then demand a proof, just as we demand of you. > > To wit: > > Let x be any natural. Then 2x is divisible by 2 and is a > member of the set of evens. Thus f:N->E > > Let y be any even, i.e. y = 2x for some natural. Therefore > f(x) = y, i.e., every even is the image of some natural > number. Thus f is surjective. > > Let f(x) = f(x') = y for some x, x' in N. Then 2x = 2x' = y, > or (2x-2x') = 0, i.e. x=x'. Since f(x)=f(x') => x = x', then > f is injective. > > Since f:N->E is both injective and surjective, it is bijective. Okay, thanks, how about a proof of bijection between the natural numbers and the binary strings? Could you show me that, with wording to your liking? I think the combination should do quite nicely for this bijection. Thanks in advance. > > ----------------- > > Conspicuously absent here is any TO-ish mention of a "last" or > "largest" element. By letting x or y be arbitrary elements, > the reasoning applies to every element of the respective sets. > > - Randy > > -- Smiles, Tony
From: Tony Orlow on 27 Oct 2005 11:11 William Hughes said: > > Tony Orlow wrote: > > William Hughes said: > > > > > > Tony Orlow wrote: > > > > David R Tribble said: > > > > > Tony Orlow wrote: > > > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > > > > > > > Virgil said: > > > > > >> {x in S:x not in f(x)} > > > > > > > > > > > > > > > > Tony Orlow wrote: > > > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > > > > > Your mapping does not include any natural that maps to the entire set > > > > > *N, which it must do in order to be called a bijection, since *N is one > > > > > of the members of P(*N). Show us that one single mapping, please. > > > > It would obviously be the natural with an infinite unending strings of 1's, > > > > right? Ah, but you want to know, how MANY 1's? > > > > > > No, this is irrelevent. No natural has a binary representation that > > > consists only of 1's. > > I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit > > the bill. Why do you say I cannot declare that the 1's go on indefinitely? We > > are talking about *N, the set including infinite naturals, with infinite > > numbers of bits. When will I run out of bits for my subset elements? Remember, > > there is no last element, as you remind me again, below. > > > You do not "run out of bits". > The reason that no natural number can > have a binary representation composed only of ones > is that every natural has a successor which also > has a binary representation. > > Suppose that a natural n > had a binary representation that is composed only of ones. > What could the representation of n+1 be? It cannot consist > only of ones (this is the binary representation of n). It cannot > have a 0, as a binary representation containing a 0 is the binary > representation of a number smaller than the number with a binary > representation consisting only of ones. Thus n+1 cannot have a binary > representation. Contradiction. Well, golly gee. A contradiction. And all you asked me for was a natural with a 1 bit for every natural in the set. What did you expect? Of course there are problems with the answer. There are problems with the question. That's why you get a contradiction. You want the complete mapping to every element of an endless set. I want a flying pink unicorn that pees single-malt scotch. Oh well. Guess we're both SOL. > > -William Hughes > > -- Smiles, Tony
From: albstorz on 27 Oct 2005 11:11
David Kastrup wrote: > Albrecht wrote: > > You misinterpret totally when you say, I think there must be an > > infinite natural number. I don't think so. I only argue that, if there > > are infinite sets, there must be infinite natural numbers (since nat. > > numbers are sets). > > That is like saying if there are animals in a zoo, there must be an > elephant (since an elephant is an animal). Too bad you'll find zoos > without an elephant in them. Wrong. That's like saying if a zoo containes animals, there must be animals in the zoo. > > > I don't say: there are infinite sets. You say: there are infinite > > sets and there is no infinite number. > > There is a zoo without an elephant in them (why, my home town has > one). There is a zoo without animals. > > > And I say: If there are infinite sets there must be infinite > > numbers. > > You say: if there is a zoo, there must be an elephant. If there is no animal, there is no zoo. .... Regards AS |