traversing an infinite collection

Discussion in 'Physics & Math' started by chinglu, Aug 8, 2016.

Thread Status:
Not open for further replies.
  1. chinglu Valued Senior Member

    Messages:
    1,637
    Say you have a countable infinite sequence A that is contained in some interval [a,b].

    Prove a clock can tick through all elements of A when time elapses from a to b.

    RPenner, show your "math skills".
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. The God Valued Senior Member

    Messages:
    3,546
    Countable infinite sequence !!
    This term does not belong to maths.
    My clock ticks everything between (12,12), nothing more nothing less. So disproved for any (a,b).
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Fednis48 Registered Senior Member

    Messages:
    725
    Your question is ill-posed. A clock does not "tick through elements of a sequence" - its hands rotate in 3D space. Of course you can map the hand positions to elements of a sequence, but you'd need to be a lot more precise about it to form a good question. If nothing else, the quantum uncertainties in the hand positions mean you'll have to clarify what it means for a hand to "point at" a particular element.

    On a related note, I think you'll learn a lot more if you drop your crusade to discredit the best mathematician on this site.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. exchemist Valued Senior Member

    Messages:
    12,453
  8. mathman Valued Senior Member

    Messages:
    2,002
    A physical clock can't. If you have a clock that can keep halving the space between ticks, you could use that.
     
  9. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    It's not clear if the intended topic of discussion is
    1) Physical clocks doing some sort of discrete physical operation called "ticking"
    2) The assumption that the nature of time is continuous
    3) A theorem that an arbitrary countably infinite partition of a finite interval can be summed to have the same value as that interval.
    4) Something else.

    Pure physics: If we are talking about some sort of discrete physical operation called "ticking" and each tick must proceed serially and each tick takes a finite minimum time, then of course there is a finite upper limit to how many ticks you can have have in a finite interval of time.

    Pure math: If we are talking about various ways to sum the infinite partition of a finite interval, then since the partition admits an order of its elements based on descending size, and position within the interval, then it follows that the ordered partition is absolutely convergent. Therefore the ordered partition is universally convergent. In fact the Dvoretzky-Rogers theorem says that for a finite-dimensional Banach space, like the 1-dimensional real number line, absolute convergence and universal convergence go hand-in-hand. Therefore the partition can be summed in any order to get the same limit, therefore the ticks in the order given add up to be the whole interval.

    In between those two extremes, we have a mishmash of physics and math notions with no guidance from the OP what the topic is.
     
    Yazata likes this.
  10. The God Valued Senior Member

    Messages:
    3,546
    Read along with the context. You cannot count the infinite sequence in finite steps...try that standard infinite sequence 1+1/2+1/4+1/8+1/16..........it can never reach at 2 in finite counts/steps.
     
  11. QuarkHead Remedial Math Student Valued Senior Member

    Messages:
    1,740
    This is not a sequence, it is a series. So Bzzzzt!

    Suppose the interval \([a,b]\) is uncountable, with a countable but non finite subset \(A\). This means, by definition, there is a bijection from \(\mathbb{N}\), the natural numbers, to \(A\). One can call this a "sequence" because a sequence is defined as an injective mapping \(S:\mathbb{N} \to [a,b]\) such that \(S(n) \equiv x_n\) for any \(n \in \mathbb{N}\) and some \(x \in [a,b]\)

    Suppose that the time interval \([t_1,t_2]\) is countable but not finite, that is "time is infinite but not continuous". Then, by the above, there is nothing to prove.

    Suppose on the contrary that "time is continuous". Then assuming this implies (why should we?) that \([t_1,t_2]\) is uncountable, all we need do is notice that every set has a countable subset (Proof: start counting and stop when the set is exhausted or Hell freezes over, whichever is the sooner).
     
  12. chinglu Valued Senior Member

    Messages:
    1,637
    does not work
     
  13. chinglu Valued Senior Member

    Messages:
    1,637
    theoretical clock that functions exactly as time does.

    it's an abstraction.

    anyway can you prove the sequence can be traversed?
     
  14. chinglu Valued Senior Member

    Messages:
    1,637
    yes, OK, it is an abstract clock that conceptually function as time does.
     
  15. chinglu Valued Senior Member

    Messages:
    1,637
    this is not a valid proof in current mathematics.

    to prove an infinite set has a countable infinite subset you must employ the axiom of choice.

    this can be done recursively.
    case 0: pick a0 in A
    case n+1: pick an+1 in A - {a0,....,an}
     
    Last edited: Aug 10, 2016
  16. someguy1 Registered Senior Member

    Messages:
    727
    Proposed counterexample. Consider \([0,1]\) with \(a_n = \frac{1}{2} + \frac{1}{n}\) for \(n = 2, 3, 4\dots\)

    The point being that time goes from $0$ to $1$ but the sequence goes backward, making it impossible to satisfy your "tick" condition.

    I say proposed because the problem is so poorly specified it's hard to nail it down. What is a clock, what is a tick, what's the nature of time? Others have noted these issues.
     
  17. QuarkHead Remedial Math Student Valued Senior Member

    Messages:
    1,740
    I am surprised that so few posters here seem to understand the difference between a sequence and a series.

    Look, a sequence is simply an ordered set. It may be finite or it may not.

    A series is the sum of infinitely many elements. This can be "converted" to a sequence by taking the partial sums in order.
     
  18. The God Valued Senior Member

    Messages:
    3,546
    I think it is more to do with not being exact.

    See, a series can have finite elements too.
     
  19. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    The opening post seems to be seeking a private conversation with user "rpenner". That is best done using the private conversation facility.

    Thread closed.
     
Thread Status:
Not open for further replies.

Share This Page