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Question about open sets in (-infinite,5]

  1. Apr 4, 2012 #1
    The stupid question of the day.

    If S is the real interval (-infinite, 5], and I can find a metric d so that (S,d) is a metric space, then,

    is, for example, (4, 5] an open set in (S,d) ?

    I say this because, the way I'm reading the definition of an open ball, the open ball B(5,1) is the interval (4,5] and not the interval (4,6), since the points in (5,6) do not belong to the metric space (S,d). So every open ball in (4,5] centered in 5 is completely contained in (4,5].
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 4, 2012 #2
    Yes, (4,5] is open.
  4. Apr 4, 2012 #3
    Thanks, micromass, just checking the fundamentals.

    By the way, I apologize for the phrasing of the question; if the metric d is not specifically defined, then there is really no way to tell. As someone else pointed me out, the question should have referred to the restriction of the Euclidean metric to the set S.
  5. Apr 5, 2012 #4
    In that case you automatically have to deal with the quotient topology on S and obviously (4,5] is certainly the intersection of an open set of R and S. Of course just using the definition of metric also works. All points with distance smaller then 1 are in B(5,1) but, this means of course all point that are in your space. Otherwise it wouldn't make much sence.
  6. Apr 12, 2012 #5
    Not necessarily. The question posed by the OP is if he can find a metric turning [itex](-\infty, 5] [/itex] into a metric space, then is (4,5] necessarily open. This is the question as it's posed, and the answer is not necessarily. He did not specify what the metric would be, so there's no guarantee that the metric would in any way resemble the regular Euclidean metric. There are many metrics that one can construct on [itex]\mathbb{R}[/itex], and then a restriction to his subspace S yields a metric on S. In some of those metrics, (4,5] may be open. In others, they may not.

    If the question were specifically about the regular Euclidean metric, then the answer is yes.
  7. Apr 13, 2012 #6
    ... read third post
  8. Apr 13, 2012 #7
    Thanks all for your answers! My actual doubt was about open balls with the Euclidean metric, but I did a awful job formulating it -- it's clear now.
  9. Apr 13, 2012 #8
    The quotient topology, as far as I learned, deals with an equivalent relation on the space and is defined on the space of equivalent classes of the original space. This case is just a subspace defined by the intersection of the open sets with the subset. Correct me if I'm wrong.
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