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A discrete subset of a metric space is open and closed

  1. Dec 10, 2012 #1

    If [itex]X \LARGE[/itex] is a metric space and [itex]E \subset X[/itex] is a discrete set then is [itex]E \LARGE[/itex] open or closed or both?

    Here's my understanding:

    [itex]E \LARGE[/itex] is closed relative to [itex]X \LARGE[/itex].
    proof: If [itex]p \subset E[/itex] then by definition [itex]p \LARGE[/itex] is an isolated point of [itex]E \LARGE[/itex], which implies that [itex]p \LARGE[/itex] is not an interior point of [itex]E\LARGE[/itex]. Since a subset of a metric space is open if every point of that subset is an interior point, it follows that [itex]E \large[/itex] is closed.

    Now that I write this proof, I believe the problem is that a subset is open IF every point is an interior point, not IF AND ONLY IF. There are other ways for the subset to be open other than all points being interior. Is this my error?

    The web tells me a discrete set is open and closed. Proofs I've seen use collections of neighborhoods with radius s.t. the only point contained in the nbhd is the discrete point at its center and since any collection of open sets is open then the discrete set is open.
    The Problem: when proving closure by taking complements, wouldn't we need ANY collection of closed sets to be closed when if fact this is only true if the collection is finite?

    Thanks for the help,
  2. jcsd
  3. Dec 10, 2012 #2
    I think you mean p \in E, not p \subset E.

    Sounds good.

    Not sure how you got that. I think you mean E is not open. That does not imply that E is closed.

    This misunderstanding I've heard of before. In defintions, like that of open, the "if" really means "if and only if". But defintions will continue to use just "if". So a set is called open if (and only if) all points are interior.

    You may not have described the problem fully. E is certainly open relative to itself, but not necessarily relative to X.

    I'm not even sure what collection of closed sets you're talking about. Please try to be more clear in your questions, it makes it easier to reply to.

    Also, be careful about your use of the word collection. I think you need to explicitly specify if you intedn union, or intersection.

    For closed sets,

    an arbitrary intersection of closed sets is closed.

    a finite union of closed sets is closed.

    an arbitrary union of closed sets is not generally closed (some are).
    Last edited: Dec 10, 2012
  4. Dec 11, 2012 #3
    yes, my mistake, [itex]p \in E[/itex]

    Here's how: every point in E is an isolated point => E has no limit points => E contains all of its limit points => E closed. (As far as I can tell, this is true for a discrete set containing finitely many or infinitely many points)
    I was wrong to say not open implies closed, which I know realize is false. I should have gone straight to the definition of closed as I have now done. Is this now correct?

    I apologize for not being exact. Let me just state my question: If [itex]\small X[/itex] is a metric space and [itex]\small E[/itex] is a discrete subset of [itex]\small X[/itex], is [itex]\small E[/itex] open, closed, or both? [itex]\small E[/itex] can be either finite or infinite.

    Thanks for your help
  5. Dec 11, 2012 #4
    And I mean is E open/closed/both with respect to X
  6. Dec 11, 2012 #5
    I think you've decided E is closed. I agree. I can't confirm your proof however, as I can't stand trying to remember or look up all those derived set, limit points, adherent point defintions.

    You've got a great attitude, don't worry about being too exact on a forum, I'm being a crank. I'm imprecise on these things quite often.

    My intuition tells me E is not necessarily open, have you decided, can you picture it?

    And you're very welcome.
  7. Dec 11, 2012 #6
    Rudin states "E is open if every point of E is an interior point of E", so if this "if" is in fact IF AND ONLY IF, then a discrete set cannot be open b/c all points in E are isolated.
  8. Dec 11, 2012 #7
    So you are claiming that isolated points cannot be interior? How would you prove this?
  9. Dec 11, 2012 #8
    Claim: isolated points cannot be interior points of any discrete subset E of a metric space

    proof: by definition, all interior points are limit points. If p is an isolated point of E, we can find a nbhd N(p) such that p is the only point in N(p). Therefore p cannot be a limit point of E and hence cannot be an interior point of E.
  10. Dec 11, 2012 #9
    How would you prove this?
  11. Dec 11, 2012 #10
    I'm actually wrong. The statement - all interior points are limit points - is not true in general, namely when E is a discrete set it is not true!! If E = [0,1] then yes every interior point is a limit point.
    proof: if p is an interior point then there exists a neighborhood N s.t. N is a subset of E. Consider all neighborhoods which are subsets of N. Each one of these neighborhoods contains a point q not equal to p s.t. q is in E, because for every radius r>0, we can find a q in E s.t. d(p,q)<r for all p in E. It is obvious that any neighborhood with radius greater than N satisfies the requirement of a limit point of E. Therefore every interior point of E is a limit point of E.
    (Any improvements are welcome)

    What I did not realize was that by definition every subset (singleton) of a discrete set is an open set. However, I don't know why this is the case.
  12. Dec 11, 2012 #11
    You are wrong, find the mistake.

    Hmm. You need to slow down and specify what X is.

    You seemed to have two different claims before you started your proof, so I must confess I am not going to take the time to read this proof, since I don't know what your claim is.

    This is not true in general.

    Are you taking a class. You may want to find better feedback through other resources, I personally do not want to continute to check every one of your sentences. You should be checking more of your own logic before I would continue to comment. You might attempt to carefully write a claim, write it's proof, then step away, come back to it a little later, and check each definition, and each step in your logic.

    When you carefully state your claim (for instance to your self), including writing it down, then develop your proof, you may find, if you're really careful, that you are having difficulty proving it. With the investigation you have just developed, you may be able to modify your claim, to a better "guess". Lather rinse repeat.
  13. Dec 11, 2012 #12
    I appreciate the feedback. Yeah I'm taking a class and unfortunately I'm having to rush through the material
  14. Dec 11, 2012 #13
    No sweat, best of luck. Yeah, having to rush through math is not fun, it's sort of like the squirrel who saves nuts for the winter is well off, school is much more of a pain when we're rushing.

    But I bet you're enjoying it, and don't let my impulsive crankiness stop you from thinking out loud on here.
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