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Homework Help: Baby Rudin, Chapter 2, Problem #20.

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Are closures and interiors of connected sets always connected?



    2. Relevant equations



    3. The attempt at a solution

    I think this is true for closures and I know it's false for interiors (Found a Counterexample.) I'm having trouble proving it for closures though; I've tried a arguments along the following lines...

    Assume X is connected but X(closure) is not. Then X(closure) = AuB, for some open separated sets A and B. So, x in X(closure), then x in A or B. x is in X or a limit point of X; so if I could show x cannot be a limit point of x, then X=AuB and I'd have the result I want. So assume x in X'. Then all N(x) contain some y in X. x is also in A or B, and A and B are open, so lets say x in A. Then x is an interior point of A; and some N(x) is contained in A. We therefore know that X intersect A is not empty. So X(closure) intersect A(closure) is also not empty. But this is not a useful result; it just says A(closure) intersect (AuB) is not empty... lol. I don't think this will work, because there is no reason (I can think of), that A and X can't intersect (which is what this contradiction would depend on.), and I also see no reason a limit point of X couldn't be an interior point of A, assuming that they can intersect... sad ; (. (I later tried a small modification of this to try and show that B was empty... but it also failed miserably)

    So... now I'm trying to show from A intersect B(closure) = Null and B intersect A(closure = Null; that X must be the null set. But how else can I express X(closure) so that I can work with it and extract information about X from it, other than as XuX'?

    Intuitively, I'm feeling this is probably true because if X is connected, then "adding" the "boundary" points (roughly speaking) isn't going to make it "unconnected." So I feel like I should be able to derive the contradiction, "If X(closure) is not connected, then X is not connected, but we said it was!) However I am just awful at proving things, or am not seeing the trick here.

    Ugh. The adjustment to proof based math is not treating me well...
     
  2. jcsd
  3. Dec 14, 2008 #2
    Suppose U and V are two disjoint, open sets and assume cl(X) is contained in the union of U and V (so X is automatically in the union). Since X is connected, we know that it lies entirely in U (say). Now consider the complement of V.
     
  4. Dec 14, 2008 #3
    Hm... ok... so...

    Assume that X is connected, but X(closure) is not connected. Then X(closure) is contained in AuB, and X is also contained in AuB; Since is connected, X must be contained in either A or B, else it'd be the union of two non-empty, separated, open sets and we'd have a contradiction. So assume X is contained in A. Then X(intersect)B = Null. B(complement) = A(closure), since B is open. But that means X(closure) is contained in A (or else we could have some limit point of X not in A, and A would not be closed). This means X closure is contained in A (Because the intersection X(closure), B must be Null), so X closure is not the Union of two non-empty open sets(B must be empty)(?); X closure is connected; a contradiction.

    That works I think! But why does X(closure) being in A imply X(closure) is connected? It implies it's a closed subset of an open set; couldn't we have two open subsets of A, U and V which contain X closure? It can't be generally true that no subset of an open set is connected... I was hoping to run into the contradiction, If X(closure) not connected, "then X is not connected, but we said it was" Logically equivalent I guess.
     
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