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Posers and Puzzles

Posers and Puzzles

  1. 08 Nov '04 19:13
    For you math people out here: someone told me the following suspicion:

    There exists an overcountable subset of R that isn't dense in R.
    (overcountable: not isomorphic/no bijection with N; dense: every interval* around every element (I do not mean "all elements" here) of the set includes more elements from that set)

    I have the idea that it isn't true, because Q is countable and already dense in R. Can anyone shed some light on this?

    *: I'm not sure this is the correct word, but you get the idea: (a,b) \subset R with a,b \in R, a<b.
  2. Donation Acolyte
    Now With Added BA
    10 Nov '04 10:30 / 1 edit
    How about [0,1]u{2}

    Seriously though, there are uncountable sets in R which are nowhere dense, ie they're not dense within any given interval. For example, take the set of all real numbers such that the decimal expansion of each consists entirely of 0s and 1s.
  3. 11 Nov '04 11:55
    I too late realised that I had posed the question wrong

    What I meant was that the subset should be isolated (which is indeed, dense in no interval). But it appears that this is impossible. I don't know the exact proof, but the trick is that you add the size of the intervals around the elements which include no more elements. This amounts to more than R if the subset is un/over-countable. (shown for example after mapping R onto an interval with arctan - you would get more than the length of the interval).

    BTW, The set you described is dense: take an element b, and make an interval around it with length e<1. In the decimal expansion of e/4 (to be safe ), there is a place which is the first that isn't 0. Change one of the decimals of b after that place, and you have another element in the interval. So your set is dense in R.