1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Baby Rudin, Chapter 2, Problem #20.
  1. 2 Rudin Problems (Replies: 0)

  2. Baby Rudin Problem 2.7 (Replies: 2)

  3. Baby Rudin problem (Replies: 4)