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A point-set theorem

  1. Mar 4, 2005 #1
    I am trying prove the following theorem:

    If a set [tex]A[/tex] is connected, and if [tex]A \subseteq B[/tex] and [tex]B \subseteq cl(A)[/tex] (where cl(A) is the closure of A), then [tex]B[/tex] is connected.

    Just so everyone is on the same page, a set X is connected if the only subsets of X that are both open and closed are the empty set and X itself. I gave the definition in case that isn't the most common definition. I'm sure there are a number of ways of saying it.

    I thought I had something going, and all I needed was to show that if a set is open, then the interior of its closure is the set itself (hence my last post). Of course this is false: take any bounded open interval on the real line and remove one point from the middle. Any suggestions?

    thanks
     
  2. jcsd
  3. Mar 4, 2005 #2

    Hurkyl

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    Slight correction -- the openness and closedness are relative to X.

    This version seems simpler to use:

    A set S is disconnected iff it can be decomposed into two nonempty, disjoint sets whose closures are also disjoint.
     
    Last edited: Mar 4, 2005
  4. Mar 4, 2005 #3

    I like your definition better. Two questions though:

    1) In your def., do the two disjoint sets need to be open (in some larger set X, for instance)?

    2) What exactly is meant by "relatively closed (open)"? The book I have seems to hint that such a concept exists, but since it isn't really a book on topology, it doesn't give a definition explicitly.
     
  5. Mar 4, 2005 #4

    Hurkyl

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    No, the two sets do not need to be open.

    For example, if S = [0, 1] U [2, 3], then I can decompose it into the two sets A = [0, 1] and B = [2, 3]. closure(A) and closure(B) are disjoint, thus S is disconnected.


    Given a set X, a subset Y of X is open with respect to X
    iff
    Y is the intersection of X and an open set
    iff
    Any point P of Y has a neighborhood N that does not contain any point of X \ Y.
    iff
    (insert a variant of your favorite characterization of open sets)


    Another way of describing it is to specify the relative topology. Given a subset X of a topological space, we can define a topology on X whose open sets are precisely the open sets of the whole space intersected with X.
     
    Last edited: Mar 4, 2005
  6. Mar 4, 2005 #5
    Awesome. Thanks for the explanations, Hurkyl. I will think about it some more.

    Of course, I am still accepting suggestions on how to prove the above theorem. :smile:
     
  7. Mar 4, 2005 #6

    Hurkyl

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    I usually like the contrapositive, especially when dealing with the first principles of connectedness. Assume B is disconnected, then prove A is disconnected.
     
  8. Mar 4, 2005 #7
    That was the direction I started in, but I crapped out pretty soon after. Here's what I got:

    Recap of Proposition: If A (a subset of a topological space X) is connected and B contains A and closure(A) contains B, then B is connected.

    Assume B is disconnected. Then there exist nonempty disjoint sets G and H whose union is B. Then either [tex]A \subseteq G[/tex] or [tex]A \subseteq H[/tex]. Otherwise the sets (A intersect G) and (A intersect H) would provide sets that show A is disconnected, contrary to hypothesis. So assume A is contained in G. Then closure(A) is contained in closure(G). But since closure(A) contains B, we have closure(G) contains B. Then closure(G) contains H.

    Now if the contrapositive definition of connectedness states that the closures of G and H must be disjoint, then we have reached a contradiction and that's that.

    The thing is, I'm not 100 percent convinced the closures of G and H must be disjoint. Well, at least not according to the definition that I was first using, in which a set X is connected if the only subsets of X that are both relatively open and closed are the empty set and X. From this version, one can derive the contrapositive statement:

    A set X is not connected iff there exist disjoint, relatively open (to X), nonempty subsets G and H of X such that (G union H) = X.

    That doesn't seem to require that the closures of G and H are disjoint. Is it possible that our definitions of connected aren't equivalent?
     
  9. Mar 5, 2005 #8
    As it turns out, this statement also appears to hold true if we replace "relatively open" with "relatively closed". In that case their closures (in X) would also be disjoint. Does this make our definitions compatible?
     
  10. Mar 5, 2005 #9

    mathwonk

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    here is a very powerful version of connectedness that is nonetheless trivially equivalent to the others: a set S is connected if and only if every continuous map from S to the two point set {0,1} (with usual discrete topology) is constant.

    when combined with any of the usual properties of a continuous function, this version makes the problem above fairly trivial.
     
  11. Mar 6, 2005 #10

    Thanks for the definition!

    I have been thinking about it quite a while now. I showed the above definition is equivalent to the others. However, my attempts to prove the theorem were more cumbersome than my original attempt, and did not work.

    Could you give me a hint that will help me see how it is trivial? :biggrin:

    thank you
     
  12. Mar 6, 2005 #11

    Hurkyl

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    How does [itex]\lim f(x)[/itex] compare to [itex]f(\lim x)[/itex]?
     
  13. Mar 7, 2005 #12

    They are equal everywhere f is defined, right?


    Here is what I think you are trying to tell me:

    Assume B is disconnected. Then there is a continuous onto function f: B->{0, 1}. Define a function g from closure(A) to {0, 1} so that for an element y of B, g(y) = f(y). For elements z not in B, let g(z) = lim f(x) as x->z. Then g is a continuous onto function from closure(A) to {0, 1}, and closure(A) is disconnected. But since every continuous function from A to {0, 1} is constant, the same must hold true for closure(A). This is a contradiction.

    Is this right? I am still working on convincing myself that the various definitions of limit and continuity are all equivalent, so for me the above argument is fuzzy.

    thanks for your help.
     
  14. Mar 7, 2005 #13

    Hurkyl

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    The problem is that you have not proven that every function from closure(A) -> {0, 1} is constant, just the ones you've constructed.

    (It so happens that you've constructed every such function, but you never deduced that, nor used it in your proof)

    Incidentally, with mathwonk's definition, I think direct proofs would be more straightforward. In fact, I think what you've written is a direct proof, just wrapped up in proof by contradiction machinery.
     
  15. Mar 7, 2005 #14
    Oh, I see what you mean.

    Since A is connected, every continuous function f : A -> {0, 1} is constant. It must be the case that every continuous function g : closure(A) -> {0, 1} is constant, since for continuous functions we must have g(lim x) = lim g(x), and since closure(A) consists of (all) limit points of A (in other words, a function that was not constant would somewhere violate the condition g(lim x) = lim g(x)). But the set B also consists entirely of limit points of A, so by the same reasoning, every continuous function h : B -> {0, 1} is constant, and therefore B is connected.

    Is this it?
     
    Last edited: Mar 7, 2005
  16. Mar 7, 2005 #15

    Hurkyl

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    Yes, that's the key idea.
     
  17. Mar 7, 2005 #16
    Thank you for the help, Hurkyl and mathwonk! I got a good intro to point-set theory by thinking about this problem.

    Out of curiosity, what would be an example of a "nontrivial" theorem from point set topology?
     
  18. Mar 8, 2005 #17

    mathwonk

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    i had the following argument in mind: the inverse image of a closed set is closed under a continuous function, so if f is constant on some set it is also constant on its closure. done.

    To me, a non trivial theorem from point set topology would be the jordan curve theorem that any injective continuous map from the circle to the plane has as image a set that separates the plane into exactly two connected components.
     
  19. Mar 8, 2005 #18
    Okay. I see how that is true provided the continuous function is defined on the topological space X containing the set A. But since the function need not be constant on all of X (but at least on A), how do we know that a function continuous on A is necessarily continuous on X? The way I understood it, the continuous functions mentioned in the condition for connectedness need only be continuous on A. I am missing something obvious, I'm sure. I just don't know what.

    Oh yeah, I've heard of that! Apparently Jordan himself didn't quite prove it. Must be hard.
     
  20. Mar 8, 2005 #19

    mathwonk

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    ok. A is contained in B and B is contained in the closure of A. to prove that B is connected, let f:B-->{0,1} be any continuous map. we claim it is constant.

    At least its restriction to A is constant because A is connected. Now relative to B, the closure of A is B itself. (Closure is a relative concept.) So take B to be the whole space of interest.

    then f is constant on the closure of A in B, namely on B. hence f is constant on B.
     
    Last edited: Mar 8, 2005
  21. Mar 8, 2005 #20

    mathwonk

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    actually there are not a lot of non trivial theorems in point set topology that occur to me, until you get into really hard ones like the jordan theorem. maybe the proof that a product of compact spaces is compact is not so easy. you might try that. probably start by proving a product of connected spaces is connected.

    but a book like Kelley's General Topology, is mostly just trivial stuff you could almost prove as exercises after reading the definitions. In fact some teachers teach it that way, as a long set of exercises. This is called the "Moore method", after R.L.Moore.
     
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