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Definition of closed sets

  1. Oct 3, 2012 #1
    Good day!

    Im currently reading the book of Steven R. Lay's "Analysis with an Introduction to Proof, 3rd ed.". According to his book, if a subset S of ℝ contains all of its boundary then it is closed. But i find this wrong since if we consider S={xεQ;0≤x≤2}, then it can be shown that S contains all of its boundary points (using the fact the Q is dense in ℝ), but it is not closed since the closure of S is the interval [0,2] which is not equal to the set itself. am i correct?
     
  2. jcsd
  3. Oct 3, 2012 #2
    Is √2 a boundry point of S? Is it in S?
     
    Last edited: Oct 3, 2012
  4. Oct 3, 2012 #3

    HallsofIvy

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    Because, as you say, Q is dense in the real numbers, every irrational number between 0 and 2 (in fact, every number in S as well) is a boundary point, not just 0 and 2. S is NOT closed because it does not contain the irrational numbers. The closure of S is the interval [0, 2] including all rational and irrational numbers in that interval.
     
    Last edited: Oct 3, 2012
  5. Oct 8, 2012 #4
    May not be relevant, but you should also check which "space" you are in. In the space Q, the closure of S is S. In R, the closure of S is [0,1].
     
  6. Oct 8, 2012 #5
    Elaborating on algebrat's response, "closed subset" is a relative concept, depending on what topological space that subset is embedded in. (Obviously "open subset" is relative to the larger space as well.) This is in contrast to a property like compactness, which is intrinsic.
     
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