Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Robert Kolker on 26 Jul 2005 05:03 Han de Bruijn wrote: > > True. That's why: > > A little bit of Physics would be NO Idleness in Mathematics You are pissing and moaning that a peach is not a pear. Mathematics is deductive. Physics is empirical. Mathmematics is about the relation of ideas. Physicis is about how the real world works. Two completely different things. Bob Kolker
From: malbrain on 26 Jul 2005 07:25 stephen(a)nomail.com wrote: > In sci.math malbrain(a)yahoo.com wrote: > > stephen(a)nomail.com wrote: > >> In sci.math malbrain(a)yahoo.com wrote: > >> > Barb Knox wrote: > >> >> In article <1122338688.718048.162860(a)g47g2000cwa.googlegroups.com>, > >> >> malbrain(a)yahoo.com wrote: > >> >> > >> >> >Barb Knox wrote: > >> >> >> In article <MPG.1d4ecd45545679a8989f6b(a)newsstand.cit.cornell.edu>, > >> >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >> >> >> [snip] > >> >> >> > >> >> >> >keep in mind that > >> >> >> >inductive proof IS an infinite loop, so that incrementing in the loop > >> >> >> >creates > >> >> >> >infinite values, and the quality of finiteness is not maintained over those > >> >> >> >infinite iterations of the loop. > >> >> >> > >> >> >> Using your computational view, consider the following infinite loop > >> >> >> (using some unbounded-precision arithmetic system similar to > >> >> >> java.math.BigInteger): > >> >> >> > >> >> >> for (i = 0; ; i++) { > >> >> >> println(i); > >> >> >> } > >> >> >> > >> >> >> Now, although this is an INFINITE loop, every value printed will be > >> >> >> FINITE. Right? > >> >> > > >> >> >Not so fast. The behaviour of incrementing i after it reaches INT_MAX > >> >> >is undefined. > >> >> > >> >> Not so fast yourself. You missed the part about "unbounded-precision > >> >> arithmetic system", which has no max. > >> > >> > Sorry, but the C standard admits no such system. INT_MAX must be > >> > declared by the implementation. There is no room for exceptions. karl > >> > m > >> > >> Why are you talking about C? > > > Because C is "better" defined than java, and the example is written in > > C. karl m > > The example is not written in C. It is perfectly legal > Java code, and C++ code, and C# code, and Javascript. > Given that 'println' is not a standard C function, but > it is a standard Java function, and the author identified > the example as Java, it is pretty clear the example was written > in Java. I really don't want to get into a discussion of why the C standard doesn't allow EXCEPTIONS when considering the infinite. Perhaps later this afternoon the sky will open up a little. karl m
From: Dik T. Winter on 26 Jul 2005 07:34 In article <1122347583.518181.245300(a)g14g2000cwa.googlegroups.com> malbrain(a)yahoo.com writes: > stephen(a)nomail.com wrote: > > In sci.math malbrain(a)yahoo.com wrote: > > > Barb Knox wrote: > > >> In article <MPG.1d4ecd45545679a8989f6b(a)newsstand.cit.cornell.edu>, > > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > >> [snip] > > >> > > >> >keep in mind that > > >> >inductive proof IS an infinite loop, so that incrementing in the loop creates > > >> >infinite values, and the quality of finiteness is not maintained over those > > >> >infinite iterations of the loop. > > >> > > >> Using your computational view, consider the following infinite loop > > >> (using some unbounded-precision arithmetic system similar to > > >> java.math.BigInteger): > > >> > > >> for (i = 0; ; i++) { > > >> println(i); > > >> } > > >> > > >> Now, although this is an INFINITE loop, every value printed will be > > >> FINITE. Right? > > > > > Not so fast. The behaviour of incrementing i after it reaches INT_MAX > > > is undefined. > > > > There is no INT_MAX for an unbounded-precision arithmetic system. > > The C language is defined by the C standard, as defined by ISO. There > are no "unbounded" standard types in the C language. karl m Who is talking about C? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Barb Knox on 26 Jul 2005 07:41 In article <MPG.1d4f01ca1b87261c989f76(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: [snip] >In this case, there is justification for the axiom of induction. >Peano didn't make it up randomly. It is a statement about how a recursive >proof can cover an infinite set without requiring an infinite statement. >As such, it eally is an axiom about an infinite recursive process, >rand if we keep this in mind, and note that our inductive step >(f(n)->f(n+1)) involves an increment (n+1 is finite), > then we can see that the infinite loop that this statement >represents ends up incrementing the value an infinite number of times, >to produce an infinite value. Even in your own computational terms, that is clearly wrong. Consider the following Java infinite loop, which uses unbounded-precision arithmetic: import java.math.BigInteger; public void main(String args[]) { for (BigInteger i = BigInteger.ZERO; ; i.add(BigInteger.ONE)) { System.out.println(i); } } Now, although this is an INFINITE loop, every value printed will be FINITE. Right? So, can you now see that the INFINITE set of natural numbers has only FINITE elements? [snip] -- --------------------------- | BBB b \ Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | -----------------------------
From: malbrain on 26 Jul 2005 07:48
Barb Knox wrote: > In article <MPG.1d4f01ca1b87261c989f76(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > [snip] > > >In this case, there is justification for the axiom of induction. > >Peano didn't make it up randomly. It is a statement about how a recursive > >proof can cover an infinite set without requiring an infinite statement. > >As such, it eally is an axiom about an infinite recursive process, > >rand if we keep this in mind, and note that our inductive step > >(f(n)->f(n+1)) involves an increment (n+1 is finite), > > then we can see that the infinite loop that this statement > >represents ends up incrementing the value an infinite number of times, > >to produce an infinite value. > > Even in your own computational terms, that is clearly wrong. Consider > the following Java infinite loop, which uses unbounded-precision > arithmetic: > > import java.math.BigInteger; > public void main(String args[]) { > for (BigInteger i = BigInteger.ZERO; ; i.add(BigInteger.ONE)) { > System.out.println(i); > } > } > > Now, although this is an INFINITE loop, every value printed will be > FINITE. Right? > > So, can you now see that the INFINITE set of natural numbers has only > FINITE elements? You still forgot to cover the giant lightning bolt that comes out of the sky and THROWS AN EXCEPTION right out of your "BigInteger" package. karl m |