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Math Help - Images of a boundary of a set equal to the boundary of images of a set

  1. #1
    Senior Member Pinkk's Avatar
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    Images of a boundary of a set equal to the boundary of images of a set

    Let U and V be open sets in \mathbb{R}^{n} and let f be a one-to-one mapping from U onto V (so that there is an inverse mapping f^{-1}: V \to U). Suppose that f and f^{-1} are both continuous. Show that for any set S whose closure is contained in U we have f(\partial S) = \partial (f(S)).

    I am completely lost. I am assuming that the way to go about this is to show that one is a subset of the other but other than that I have no idea how to proceed.

    Edit: So here is my so far very vague attempt or at least idea of how this proof might lead to: If y\in f(\partial S), then f^{-1}(y)\in \partial S so for all r > 0, B(r,f^{-1}(y))\cap S \ne \emptyset and B(r, f^{-1}(y))\cap S^{c} \ne \emptyset. If this is the correct start to the proof, I am guessing that the continuity of the mappings comes into play but again, I'm lost.
    Last edited by Pinkk; September 1st 2010 at 07:03 PM.
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  2. #2
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    Hint: x\in S is interior if and only if f(x)\in f(S) is interior.
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  3. #3
    Senior Member Pinkk's Avatar
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    I assume you mean interior to their respective sets of S and f(S). So here is my attempt at showing that f(\partial S) \subset \partial f(S):

    Let y\in f(\partial S). Then f^{-1}(y)\in \partial S. So f^{-1}(y) is not in the interior of S, so y\in \partial f(S).

    Could you prove your hint statement and clarify it a bit more? I'm still confused and I don't think my attempt is right or at the very least it is not rigorous.
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  4. #4
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    Assume x\in W_1\subset S with W_1 open then f(x) \in f(W_1) \subset f(S) and, since f is open (because f^{-1} is continous) we get that f(x) is interior if x is. Reverse the roles and it's done. This, used with the fact that Int(A) \cup \partial A = cl(A) give the result.
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  5. #5
    Senior Member Pinkk's Avatar
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    I assume you mean "...since f(S) is open..." and if that is case then what we are using the fact that for any function f: \mathbb{R}^{n} \to \mathbb{R}^{m}, f continuous and U \subset \mathbb{R}^{m} open implies f^{-1}(U) open as well, correct? And so for the main proof how would we proceed? I think I have a general idea but nothing I can put down into rigorous proof.
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  6. #6
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    Well, the theorem I'm using is: If f bijective and continous then they're equivalent f^{-1} cont. and f open.

    My first hint gives f(Int(S))=Int(f(S)) and by yet another characterization fo continuity we have f(cl(S))=cl(f(S)). Can you finish from here?
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  7. #7
    Senior Member Pinkk's Avatar
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    I am just not sure what you mean by " f open" since f is a function, not a set. But I think I have some idea for a proof:

    Let x\in cl(S) such that f(x)\in f(\partial S). So f(x)\in f(cl(S)) and x\in \partial S. So x is not an interior point of S. Since f^{-1} is a continuous function, we have that f(x) is not an interior point of [tex]f(S). But f(x)\in (f(cl(S)), so f(x)\inso f(x)\in \partial (f(S)).

    I don't see why f(cl(S)) = cl(f(S)). That seems like just a different version of what I'm currently trying to prove. Sorry, I'm just having a hard time understanding this material.
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  8. #8
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    A function f is said to be open if it takes open sets to open sets. For the equality f(cl(S))=cl(f(S)) just use the definition of continuity which says: f is continous iff f^{-1}(V) is closed whenever V is.
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  9. #9
    Senior Member Pinkk's Avatar
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    I understand the definition of continuity but I just don't see how that leads to the equality statement. I think I'm just in way over my head and not ready for this type of material...
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  10. #10
    Senior Member Pinkk's Avatar
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    Okay, so I know how to prove f(cl(S)) = cl(f(S)).

    Let f(x)\in f(cl(S)). Then x\in S or x\in \partial S. If x\in S, then f(x) \in f(S), which implies f(x)\in cl(f(S)). If x\in \partial S, then for any \epsilon > 0, B(\epsilon , x) \cap S \ne \emptyset. So there is a sequence of points in S that converges to x. Since f is continuous on S, there is a sequence of points in f(S) that converges to f(x). Thus f(x) \in cl(f(S)). So in either case, f(x)\in f(cl(S)) \implies f(x)\in cl(f(S)), so f(cl(S)) \subset cl(f(S)). A similar argument leads to the conclusion that cl(f(S)) = f(cl(S)).

    But now how do I use this to prove \partial f(S) = f(\partial S)?


    Edit: Alright, here is my attempt:

    Lemma: If f is a bijection, then f(S^{c}) = (f(S))^{c}.
    Proof: Suppose to the contrary that f(S^{c}) \ne (f(S))^{c}. Then there exists y such that y\in f(S^{c})\cap f(S). So there are at least two distinct elements in the domain of f, say x_{1}, x_{2}, such that f(x_{1}) = f(x_{2}). Then f is not injective, which is a contradiction. Q.E.D.

    So now let f(a)\in f(\partial S). Then a\in \partial S, which means there is a sequence of points in S and S^{c} that converge to a, and since f is continuous, there is a sequence of points in f(S) and f(S^{c}) = (f(S))^{c} that converge to f(a), so f(a)\in cl(f(S))\cap cl((f(S))^{c}) = \partial f(S), so f(\partial S) \subset \partial f(S). A similar argument shows \partial f(S) \subset f(\partial S), so \partial f(S) = f(\partial S). Q.E.D.
    Last edited by Pinkk; September 3rd 2010 at 08:12 PM.
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  11. #11
    MHF Contributor Bruno J.'s Avatar
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    Your proof is correct, though in my opinion somewhat cumbersome. Here is a way it could be improved (in my opinion).

    Since both f and f^{-1} are continuous, f induces two bijections : (1) a bijection between the closed sets C with S\subset C \subset U and the closed sets C' with f(S) \subset C' \subset V, and (2) a bijection between the open sets O contained in S and the open sets O' contained in f(S). One way to characterize the closure \overline A of a set A is as the intersection of all closed sets containing A, and one way to characterize the interior \mbox{int }A is as the union of all open sets contained in A. So we have

    \displaystyle\overline{f(S)} = \bigcap_{C \supset f(S)} C = \bigcap_{C' \supset S} f(C') = f(\overline{S}).

    \displaystyle\mbox{int }f(S) = \bigcup_{O \subset f(S)}O = \bigcup_{O' \subset S}f(O') = f(\mbox{int }S).

    So we have \partial f(S) = \overline{f(S)} - \mbox{int }f(S) = f(\overline{S}) - f(\mbox{int }S) = f(\overline{S}-\mbox{int }S)=  f(\partial S).
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  12. #12
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    f is a homeomorphism between the topological spaces U and V. Homeomorphisms preserve all topological properties.
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  13. #13
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Iondor View Post
    f is a homeomorphism between the topological spaces U and V. Homeomorphisms preserve all topological properties.
    How would you define "topological property", other than by "property which is preserved by homeomorphisms"?
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    Quote Originally Posted by Bruno J. View Post
    How would you define "topological property", other than by "property which is preserved by homeomorphisms"?
    A property that is definable by using only the predicate "is an open set", the relation "is an element of" and all the set operations arbitrary union, complement and arbitrary intersection, the subset relation etc.
    The definition may not refer to any other representation dependent structure besides that.
    A homeomorphism preserves these predicates and therefore all properties that are definable with these predicates.

    For example the boundary of a set is equivalent to this definition
    X is the boundary of S
    iff
    X is the largest set, such that
    for all open U:
    U \cap X \neq \emptyset \rightarrow U\cap S \neq \emptyset \wedge U\cap S^{c}\neq \emptyset

    One could translate this definition into a formula of predicate logic \phi(X,S), which uses only the above mentioned predicates. Homeomorphisms preserve such formulas, that means if f is homeomorphism and \phi(X,S) is true, then also \phi(f(X),f(S)) is true.
    So if X is the boundary of S, then f(X) is the boundary of f(S).
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  15. #15
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Iondor View Post
    A property that is definable by using only the predicate "is an open set", the relation "is an element of" and all the set operations arbitrary union, complement and arbitrary intersection, the subset relation etc.
    The definition may not refer to any other representation dependent structure besides that.
    A homeomorphism preserves these predicates and therefore all properties that are definable with these predicates.

    For example the boundary of a set is equivalent to this definition
    X is the boundary of S
    iff
    X is the largest set, such that
    for all open U:
    U \cap X \neq \emptyset \rightarrow U\cap S \neq \emptyset \wedge U\cap S^{c}\neq \emptyset

    One could translate this definition into a formula of predicate logic \phi(X,S), which uses only the above mentioned predicates. Homeomorphisms preserve such formulas, that means if f is homeomorphism and \phi(X,S) is true, then also \phi(f(X),f(S)) is true.
    So if X is the boundary of S, then f(X) is the boundary of f(S).
    That sounds like a very reasonable definition!
    All the books I have seen define "topological property" as a "property preserved by homeomorphisms". It's clear, as you mention, that all properties definable within the predicate logic you describe are topological properties. How would you show the reverse inclusion? (My knowledge of logic is very limited!)
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