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Interior, Closure, Complement of sets

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let (X,d) be a metric space and E is a subset of X. Prove that
    ra18.JPG
    (c means complement, E bar means the closure of E)

    2. Relevant equations
    N/A
    3. The attempt at a solution
    Let (X,d) be a metric space and B(r,x) is the open ball of radius r about x.
    Definition: Let F be a subset of X. The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
    Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
    Definition: Let F be a subset of X. F is called closed iff whenever (xn) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F) The closure of F is the set of all limit points of sequences in F.
    Theorem: x is in the closure of F iff for all r>0, B(r,x) intersect F is nonempty.

    I wrote all the relevant definitions in front of me, but I still can't figure out this one. I know that for proving equality of sets, I have to prove that A is contianed in B(x E A => x E B) and B is contianed in A (x E B => x E A). However, for this problem, I can write down the basic definitions for each direction, but I just can't figure out the connection between the two sets...

    Any help is greatly appreciated!
     
  2. jcsd
  3. Feb 21, 2010 #2
    Suppose x is in the complement of the closure of E. This is the same as saying "x is not in the closure of E." What defining property does x have (or by definition not have) according to your definitions?
    If one way is too hard, you can always prove the contrapositive instead, as it has the same truth value.
    Ie., instead of showing that "if x is in the complement of the closure of E, then x is also in the interior of the complement of E", prove instead that "if x is not in the interior of the complement of E, then x is not in the complement of the closure of E", which is the same as "if x is in the frontier of the complement of E or in the exterior of the complement of E, then x is in the closure of E."
     
  4. Feb 21, 2010 #3
    If [tex]x \in (\overline{E})^c[/tex], then [tex]x \notin \overline{E}[/tex] so there exists some r>0 such that B(r,x) does not intersect E. But to say that it does not intersect E is the same as saying that B(r,x) is contained in [itex]E^c[/itex], and this implies that x is in the interior of [itex]E^c[/itex] so [tex]x \in \textrm{int}(E^c)[/tex]. We now have [tex] (\overline{E})^c \subseteq \textrm{int}(E^c)[/tex].

    The other direction is similar. See if you can do that.
     
  5. Feb 21, 2010 #4
    Thanks, this is really helpful!

    Suppose x E int(Ec)
    => by definition of interior, there exists r > 0 such that B(r,x) is contained in Ec
    => there exists r > 0 such that B(r,x) does not intersect E

    Now how can we prove that [tex]x \in (\overline{E})^c[/tex]??
     
  6. Feb 21, 2010 #5
    Well what is the closure? According to your definition of closure, the closure of E is the set of all limit points of sequences in E (this is a terribly confusing way to define limit points by the way, but for this purpose it doesn't matter). Since you showed that x is not in E, this takes care of the limit points that are in E. Of course a sequence in E can converge to a point that is not in E. Your job is to show that x is not one of these points either. But you have a ball B(r,x) centered at x, all of whose points do not intersect E. Is it possible for a sequence in E to converge to x?
     
  7. Feb 22, 2010 #6
    Suppose x E int(Ec)
    => by definition of interior, there exists r > 0 such that B(r,x) is contained in Ec
    => there exists r > 0 such that B(r,x) does not intersect E

    From here, can we conclude from the following theorem that [tex]x \in (\overline{E})^c[/tex]?? Are we done here? Or do we need something more?
    Theorem: x is in the closure of F iff for all r>0, B(r,x) intersect F is nonempty.
     
  8. Feb 22, 2010 #7
    Fine, you can use that theorem if you want, but the fact that you simply restate the theorem, along with everything else you've already said says something about your understanding. You're done if you can actually figure out how to conclude something from the theorem.
     
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