Cantor Confusion
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From: mueckenh on 24 Jan 2007 07:07 Dave Seaman schrieb: > > N is an infinite set, but each member of N is finite. Translated: The set of all segments {1,2,3.,..,n} is infinite, but each segment {1,2,3.,..,n} is finite. Now the number does not play a role. If there is no infinite segment, then they all put together, always starting from 1 again) cannot be actually infinite (finished infinity, reached infinity) but only potentially infinite. Regards, WM
From: mueckenh on 24 Jan 2007 07:13 William Hughes schrieb: > > > The fact > > > that you can continue a path does not mean that you must continue > > > a path. > > > > > > > Whenever there is an edge leading a path, this path is there. > > Yes, there are two paths. The path that stops and the path that > goes on. No, by definition, the trees in my proof have not leaves other than at the bottom. There is no stop inside a tree. Regards, WM
From: Andy Smith on 24 Jan 2007 07:23 maginatorium(a)despammed.com writes > (snip) > >Really? binf? What is the value of binf? > I thought I had said that there wasn't a binf and that inf was not a natural number above? >Tell us more about the difference between > >x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... (A) > >and > >x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... + >binf*(1/2)^[?!@?] (B) > >Do you understand that (A) is the 'sum' of an unending series (defined >properly in terms of limits, and with not an 'inf' in sight)? It goes >on without end - there is no 'inf' digit when you get to the end, >because you can't get to the end, because there isn't one. (Gosh, I've >typed this before, I suspect). So if you insist on adding something >called 'binf', it is quite separate from the unending series, and its >function is actually [!] rather obscure. > Yes, I agree, no problem. >Now go back and rephrase your statement above, eliminating the word >"actual". Thinking clearly about whether something has an end or not >should help. > There is no last term. But in a real e.g. . 0.011... all the unending sequence of 1's are required to distinguish it from some other real. So if you systematically try to index the reals as I suggested, then you require an infinite number ...110. So you can't index the reals, at least not by my suggested systematic technique - which I think is equivalent to any systematic technique. Not an actual in sight. -- Andy Smith
From: imaginatorium on 24 Jan 2007 07:26 Andy Smith wrote: > In message <gccer2dvlhm5q04g7deea6ro0t0c5b31p6(a)4ax.com>, G. Frege > <nomail(a)invalid.?.invalid> writes > >On Wed, 24 Jan 2007 08:43:43 +0100, G. Frege <nomail(a)invalid> wrote: > > > >>> > >>> If that's what he means, then I'll agree he could be close. It isn't a > >>> proof, but as a heuristic it is OK. > >>> > >> Though by using (almost) the same heuristic he might conclude that > >> rational numbers aren't countable too. Well... > >> > >Or even better: > > > >Consider a countable subset of the set of real numbers containing only > >irrational numbers. > > > >Then the argument (the heuristic) of the OP might lead him to the > >conclusion that this subset is not countable. > > > > > >F. > > > Well we were talking about the uncountability of the reals, not > necessarily proposing a computer science perspective as a replacement > methodology for your number theory ... > > But if your subset is countable, then its definition implies that the > subset can be packed into a smaller address space (though whether it > would be easy to determine that from the definition is TBD) On what sort of basis do you assert: "can be packed into a smaller address space"? Perhaps general experience of programming? You know that if the range of values for something is greatly reduced, you save bits. But if management come and announce that they have drastically reduced the number of product codes from 8192 to 4097, and how many megabytes will you save on a squillion-widget database...? How many bits do you need to represent all natural numbers? Well, obviously no finite number is enough: you need an unending string of them. ("Unending" meaning, curiously, that there is exactly one end, but no "other" end. OK?) But exactly this same unending string of bits is enough to represent the reals between 0 and 1. Here's a picture: [ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ... This is meant to show a one-ended register: there is a first box (between [ and | ), a second box, and so on; never stopping. I know, I'll call this a registe (pronounced in mock-French), because it has no right end. We can represent natural numbers within this registe by putting the units column in the first box, the twos column in the 2nd box, and so on. No natural number will ever "fill" the registe, because every natural number has only so many bit positions; but we cannot make the registe any "shorter" without chopping it off at some point, whereupon at least some naturals will no longer fit in. Meanwhile we can also represent the reals in [0,1] by writing unending binary fractions in the registe. So any intuitions you might have about "infinite bits" and "countable sets" are undoubtedly wrong. Brian Chandler http://imaginatorium.org > -- > Andy Smith
From: stephen on 24 Jan 2007 07:27
Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > In message <1baer212phrkloe4i9r6ghiq9c451h08fm(a)4ax.com>, G. Frege > <nomail(a)invalid.?.invalid> writes >>On Wed, 24 Jan 2007 09:38:00 GMT, Andy Smith >><Andy(a)phoenixsystems.co.uk> wrote: >> >>>> >>>> To represent a real number between 0 and 1, you need 1 bit for >>>> each positive integer. >>>> >>>> x = b1*(1/2)^1 + b2*(1/2)^2 + b3*(1/2)^3 + b4*(1/2)^4 .... >>>> >>>> Why do you think this requires an "actually infinite" integer? >>>> >>> Because you need binf to complete the sum; inf is not a natural number >>> and you need an actual infinity of bits to describe it. >>> >>You are talking nonsense, again. >> >>Aren't you able to understand the difference between an >> >> (a) infinite integer >>and >> (b) infinitely many integers >>? > Yes, although my use of your terminology probably goes adrift. >> >>We do not need "binf" (?) to "complete the sum" because this sum is >>_never_ "completed". >> > Exactly so. A transcendental requires an infinite number of bits to > represent it (so as to distinguish it from the rest of the infinite set > of other real numbers). >>> >>> If you systematically try to address (map) the reals you need integers >>> with as many bits as the reals; NaN. >>> >>Bla bla. Seems that you are desperately striving for a career as a >>crank here. >> >>Go ahead, I think you will succeed! :-) >> >> >>F. >> > OK, I will take the personal abuse, probably deserved. Shouldn't put my > head above the parapet when I'm not qualified to do so. All that I was > trying to observe was that all systematic numbering schemes of the reals > are equivalent, corresponding to permutations of bit positions. And if > you adopt one systematic scheme, such as starting with bit 0, then bit > 1, etc, because the reals have an address space of an infinite number of > bits, the corresponding numbers/indices must also require an infinite > number of bits, and no natural number is infinite. So you can't do it - > you don't have a big enough address space. > Of course that may well be a load of bollocks Of course it is. There are an infinite number of integers, and an infinite number of bit positions. The address space is exactly the right size. Why do you think there are fewer integers than bit positions? Your whole argument seems based on the idea that there must be a bit oo, and that because there is no natural number oo, that it is impossible to describe a real number. But there is no bit oo. There is no last bit. There is an unending sequence of bits. Unending sequences do not end. Stephen |