Cantor Confusion
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From: David Marcus on 23 Jan 2007 19:45 Andy Smith wrote: > David Marcus writes > >> > >> Yes, but on a simplistic view of infinity (e.g. mine as of 2 weeks ago) > >> there is no reason not to think that you can have ...1111 as well as > >> .1111... > > > >How do you go from "have ...1111" to concluding that "...1111" denotes a > >natural number? We said above what a natural number is. Why do you think > >the string "...11111" identifies or names a natural number? > > > > I just meant, that on a simplistic lay view of infinity, that there is > no reason that you shouldn't have an infinite number of digits in a > number. OK, that isn't correct, but on a simple view of infinity it > seems plausible. Doesn't work with natural numbers. OK, but I rather doubt that most people would think that there are any natural numbers with an infinite number of digits. > >> - or that the Universe has to have had a beginning (everything > >> has to succeed to something previous, so there must be an origin, so you > >> can't have infinite negative time - but can conceive of an infinite > >> future time, Aristotle again. > > > >Huh? > > People cannot conceive of an infinite past, Why not? I believe that was the usual assumption before the Big Bang was discovered. > but are content with an > infinite future. There is something in common here with the natural > numbers and the Peano axioms - there are no negative numbers, all > numbers are the successor of some successor to 0. Perhaps you would > object to the idea of an infinite past on the grounds that this would > imply the existence of infinite integers (e.g. a clock started infinity > ago , but still ticking and showing an infinite integer on its dial?) Why in the world do we need infinite integers to have no beginning? The real numbers provide a fine model for time that extends indefinitely into the past. -- David Marcus
From: David Marcus on 23 Jan 2007 19:49 Andy Smith wrote: > David Marcus writes > >> > >> Sure. But (almost) all real numbers have a random infinite bit sequence, > >> and any one instance of those requires an infinite number of bits to > >> specify it (is what I meant). > > > >Why should that observation have anything to do with whether the reals > >are countable or uncountable? > > > It is just address space isn't it? You can't address 2^64 numbers with > 32 bits, and you can't address 2^(actually infinite) numbers with a > finite number of bits? Why are we limited to a finite number of bits? > OK, some numbers like the rationals and algebraic > numbers have compact representations, but they are 0% of the total ... What does "zero percent" mean in this context? > OK, that is hand waving, not a proof, but there is a clear distinction > between finite and actually infinite ...? What you wrote not only isn't a proof, but it appears to be nonsense. There is a distinction between finite and infinite, but there are infinitely many natural numbers, so it isn't at all clear that there is no bijection between the natural numbers and the reals (until you see a proof). -- David Marcus
From: stephen on 23 Jan 2007 19:45 Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: >> >>Since the integers are finite, you cannot represent a real requiring an >>actually infinite number of bits, is what I meant. Maybe that is too >>simplistic? >> >> > To be more explicit, to represent (= address) all the reals in say > [0,1] you would need as many bits for your integers as the reals occupy > . But that would require integers with an actually infinite bit length > e.g. say, the reflection of the reals about the decimal point to give > "numbers" like ...1101 Why would it require that? Can you explain your reasoning? 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? > Which is where I came in (with what's the problem with enumerating the > reals?) but I am now better informed - . If all members of N are > finite, there is no prospect of addressing the reals, so no surprise > there.... Every real number can be described using only finite bit positions. You seem intent on confusing the set with one of its elements. Stephen
From: David R Tribble on 23 Jan 2007 19:52 Andy Smith wrote: >>Since the integers are finite, you cannot represent a real requiring an >>actually infinite number of bits, is what I meant. Maybe that is too >>simplistic? >> >> To be more explicit, to represent (= address) all the reals in say >> [0,1] you would need as many bits for your integers as the reals occupy. >> But that would require integers with an actually infinite bit length >> e.g. say, the reflection of the reals about the decimal point to give >> "numbers" like ...1101 >> >> Which is where I came in (with what's the problem with enumerating the >> reals?) but I am now better informed - . If all members of N are >> finite, there is no prospect of addressing the reals, so no surprise >> there.... > Andy Smith wrote: > Are you sure you are better informed? What you wrote is nonsense. Every > real number has a binary expansion (and a decimal expansion). So, what > in the world are you trying to say? No, I think he's almost got it. I think he's trying to say (in computer programming terms) that any given real in [0,1] requires a countably infinite number of bits to represent as a binary fraction (bitstring), which is correct. And that those infinite bitstrings cannot be mapped to finite naturals (e.g., by reflecting the digits about the binary point), because you'd end up with infinite-length binary integers, which are not naturals. So I think he's reached the (correct) conclusion that you can't denumerate the reals (in [0,1]) using naturals, albeit in a somewhat clumsy way of saying it.
From: Virgil on 23 Jan 2007 20:06
In article <IMcBs3uxfqtFFwHB(a)phoenixsystems.demon.co.uk>, Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > > > >Since the integers are finite, you cannot represent a real requiring an > >actually infinite number of bits, is what I meant. Maybe that is too > >simplistic? > > > > > To be more explicit, to represent (= address) all the reals in say > [0,1] you would need as many bits for your integers as the reals occupy > . But that would require integers with an actually infinite bit length > e.g. say, the reflection of the reals about the decimal point to give > "numbers" like ...1101 > > Which is where I came in (with what's the problem with enumerating the > reals?) but I am now better informed - . If all members of N are > finite, there is no prospect of addressing the reals, so no surprise > there.... Since we have more that any finite number of natural numbers to serve as bit positions, one can easily represent all reals in the interval from 0 to 1 in any integer base from base 2 upwards. |