Cantor Confusion
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From: Franziska Neugebauer on 10 Jan 2007 12:01 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: [...] >> L_D cannot change. L_D is a line. A line cannot change. > > No. But the property of being the greatest line can and does change. Depending on the season? F. N. -- xyz
From: Franziska Neugebauer on 10 Jan 2007 12:13 mueckenh(a)rz.fh-augsburg.de wrote: [...] > The relevance is that the uninon {1,2,3,...} is *not* an infinite set If "{1, 2, 3, ...}" does not denote an "infinite set" then you should better write "{1, 2, 3, ..., n}". The trailing dots "..." are reserved for denoting sets which are infinite. > N with cardinal number aleph_0, as long as there is not an "infinite > number" *in* the set. This implication of yours waits for a proof since 2003 and has been debunked as leading to contradictions many times. F. N. -- xyz
From: Andy Smith on 10 Jan 2007 04:26 Or more easily just extend Peano to the negative numbers. Then I think you can show that -1 = infinity -2 = infinity -1 etc. (which solves my problem of a label for the last element in an infinite sequence). The infinite integers implied by negative values do not however let one index the irrational numbers ...
From: David Marcus on 10 Jan 2007 14:40 Andy Smith wrote: > > Why? You can't just stick the words "by definition" > > in your sentence to > > cover up the fact that you can't prove something. How > > you do define > > "lands on the tortoise"? The obvious way to define it > > is that the flea > > is at the same position as the tortoise. I don't see > > anything in this > > definition about there being a last jump. > > > Yes, all right, speaking too loosely. But anyway, > the flea lands on the tortoise only once, and stops. > There is a unique correspondence between landing > on the tortoise and a jump. Why? You haven't proved this. And, it isn't true. > There are no further jumps, > so landing on the tortoise is the last jump? Consider a simpler example. Suppose our location at time t is t, i.e., we move at unit speed. Then we arrive at the point 1 at time 1. What is the last time before this when we hadn't yet arrived? There isn't one. > > > What is the logical difference between this > > > So maybe, if you allow counting "infinitely fast" you can > > > count up to an infinite number? > > > > What is an "infinite number"? > > > > What if I set out on the real line, from 0 to 1, the set > 1/2,1/4,1/8, ... and then set out its mirror from 2 to 1, > moving backwards 1/2,1/4,1/8 .... > > Then the sequence {1/2+1/4+1/8 ...} + {... + 1/8 +1/4 +1/2) > covers the interval [0,2] without any gaps. > > So the infinite series {1/2+1/4+1/8 ...} + {... + 1/8 +1/4+1/2) exists > and has a finite last member = 1/2? Words in mathematics have technical meanings. A "sequence" is a function whose domain is N. So, what you wrote is not a sequence. A "series" is the sum of a sequence. So, again what you wrote is not a series. However, we can certainly partition the interval [0,2] into subintervals with the sizes you gave and in the order you gave. Then the sum of the lengths of the subintervals is 2 and the last subinterval has length 1/2. > There is undoubtedly a logical flaw here - you can't > have (by Peano) an ordered infinite series with a last member, > and doubtless you will tell me my error. Your error is your misuse of the word "sequence". > But you see the point? Not really! > Actually, at the risk of getting labelled as being in > some dodgy company, isn't there something a bit asymmetric > about Peano? What happens if we say; > > 1) All natural numbers form an ordered set bounded > by alpha (aka 0) and omega (aka infinity). 0 is a natural number. omega is not. However, if you wish to discuss ordinals and identify the finite ordinals as being the natural numbers, then it is true that omega is greater than every natural number. > 2) With the exception of alpha and omega, all > natural numbers have a successor and a predecessor; > alpha has no predecessor, omega has no successor. Correct. > 3) For any number n there exists (omega-n) which is > n. No. Subtraction is not defined for ordinals. It is defined for natural numbers, but omega is not a natural number. > Then you can happily say that there is a last jump of the flea > (but of course, the only thing you can say about the > size of the jump is that is less than any e, where e is > any real number <1. -- David Marcus
From: David Marcus on 10 Jan 2007 14:41
Andy Smith wrote: > > 1) All natural numbers form an ordered set bounded > > by alpha (aka 0) and omega (aka infinity). > > 2) With the exception of alpha and omega, all > > natural numbers have a successor and a > > d a predecessor; > > alpha has no predecessor, omega has no successor. > > 3) For any number n there exists (omega-n) which is > > > n. > > 3) obviously doesn't work (omega-n) is a number so that would give > (omega-omega+n)=n>(omega-n). omega - n is not defined, so it isn't anything. > Not so easy! > > I still want to put a sensible label on the finite last > terms of the sequence 1/2,1/4,... ...1/8,1/4,1/2 though. It isn't a sequence. However, it does have a last element. Why not call it "the last element"? -- David Marcus |