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Homework Help: Open/closed/compact sets

  1. Dec 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Can someone check this for me?
    Problem: determine which, if any of the sets if open? closed? compact?
    R=reals; Q=rationals and Z=integers.
    A= [0,1) U (1,2) is NEITHER
    B=Z is CLOSED
    C=(.5,1) U (.25,.5) U (.125, 25) U... is OPEN
    D={r*sqrt(2) such that r is an element of Q} is NEITHER
    E=R-Z is OPEN

    2. Relevant equations
    How do you find the set of accumuluation points for each of the above sets?

    3. The attempt at a solution
    my reasoning:
    A) the union is [0,1) -1. And [0,1) is neither open nor closed and 1 is bounded. so A is neither.
    B) closed by definition?
    C) union of open intervals is open
    D) Q is not closed and the complement of Q is neither (none of the points in Q are interior points so Q is not open and its complement R-Q has the same property so it is not open, therefore, Q is not closed)
    E) from (B), R-Z is the complement of Z, so it is open.
  2. jcsd
  3. Dec 14, 2009 #2
    here's what i've come up with for the accumulation points:
    C=empty set
    E=empty set.

    any help anyone?
  4. Dec 15, 2009 #3
    Your original post looks fine, but you might want to specify open/closed relative to the underlying metric space (which is probably R in each case). Also, Z closed in R isn't necessarily immediate by definition, but it's an easy thing to prove. Compactness shouldn't be hard to answer, especially if you know Heine-Borel.

    Taking the standard definition of an accumulation point of a subset A of a metric space X as a point p in which every open ball about p which contains a point q in A, q =/= p, then I would take another look at your answers. I think you may be confusing the fact that a set is closed if it contains all of its accumulation points with the false notion that an open set has no accumulation points. Also, Z is closed relative to R, but the integers cannot be the accumulation points (by the standard definition). But if a set has no accumulation points, then it is closed, because the condition is vacuously satisfied.
  5. Dec 15, 2009 #4


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    I, personally, would not use the word "neither" when there are three options.
  6. Dec 15, 2009 #5
    okay how about this:
    A=not open, not closed and therefore, not compact; accum. pt: [0,2]
    B=closed, and compact; accum pt: empty set
    C=open, and not compact; accum pt: empty set
    D=not open, not closed and therefore, not compact; accum pt: empty set
    E=open, and not compact; accum pt: empty set

    can you please check that?
  7. Dec 15, 2009 #6
    A and B look correct. However, why do you think that E for instance has no accumulation points? R-Z is just the reals with "holes" where in the integers are. If I pick a point at say 1.5, which is in R-Z, and look at any open interval about the point, I would find another real number different from 1.5 right?

    *EDIT* Also, HallsofIvy raised a point I failed to catch the first time. Ignoring compactness, you should still state whether each set is open and whether each set is closed. A set could be open and closed, and the standard example is the whole space itself (trivially satisfies the definitions). Fortunately, I'm assuming we're working with R as the underlying space, and R has the property that the subsets of R that are simultaneously open and closed are exactly the whole set and the empty set. But this should just make it easier to answer to the question of both openness and closedness.
    Last edited: Dec 15, 2009
  8. Dec 15, 2009 #7
    The integers are not compact.
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