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Math Help - Prove the image of a connected set is connected.

  1. #1
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    Prove the image of a connected set is connected.

    Problem: Let f:R^n -> R^m be continuous. Suppose A is a connected subset of R^n. Prove that f(A) is connected.

    My proof:

    Now f(A) = {f(x) : x in A}, and assume f to be an increasing function.

    Now assume to the contrary that f(A) is not connected.
    Let the two points d and e be in the set of f(A), then there exists open sets U and V in R^n such that they separate d and e.

    Now d = f(a) and e = f(b) for a and b in A. Now assume a < b.

    Since A is connected, A has Intermedian Value Property, that is, for every continuous function f, there exists a point z in A such that f(a) <= f(z) <= f(b) by the Intermedian Value Theorem.

    But that contradict our assumption as it would be impossible for U and V to separate f(A).

    Therefore we conclude f(A) is connected.

    Q.E.D.

    Yeah, I know this one is not really proper... But this is the best I could sequeeze out after almost two days of thinking...

    Thanks, please check.

    KK
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Problem: Let f:R^n -> R^m be continuous. Suppose A is a connected subset of R^n. Prove that f(A) is connected.

    My proof:

    Now f(A) = {f(x) : x in A}, and assume f to be an increasing function.

    Now assume to the contrary that f(A) is not connected.
    Let the two points d and e be in the set of f(A), then there exists open sets U and V in R^n such that they separate d and e.

    Now d = f(a) and e = f(b) for a and b in A. Now assume a < b.

    Since A is connected, A has Intermedian Value Property, that is, for every continuous function f, there exists a point z in A such that f(a) <= f(z) <= f(b) by the Intermedian Value Theorem.

    But that contradict our assumption as it would be impossible for U and V to separate f(A).

    Therefore we conclude f(A) is connected.

    Q.E.D.

    Yeah, I know this one is not really proper... But this is the best I could sequeeze out after almost two days of thinking...

    Thanks, please check.

    KK
    Sketch of proof:

    Suppose otherwise. The let a and b be points in disjoint components of f(A).

    Let C be a curve connecting a to b in A, then f(C) is in f(A), but now f(C)
    is a curve in R^m conecting f(a) to f(b), but this contradicts the assumption
    that a and b are in disconnected components of f(A).

    (here the notion of curve implies continuity, and the continuity of f ensures
    that f(C) is a curve in the appropriate sense)

    RonL
    Last edited by CaptainBlack; April 6th 2007 at 02:06 PM.
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  3. #3
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    Once again, not knowing what theorems you know it is very hard to give you a proof. In the attached pdf file I have given the standard proof that will work in any topological space. Again this may be of any use to you if you do not have the theorems used.
    Attached Files Attached Files
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  4. #4
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    So far, our class has learned:

    Defintion:

    A set B not connected if there are open sets U and V such that:

    B interset U is not empty, B interset V is not empty.
    (B interset) U U (B interset V) equals to B.
    (B interset U) interset (B interset V) is empty.

    Theorems:

    1. A is connected iff A has Intermedian Value Property.
    2. A is connected if A is pathwise connected.
    3. A is pathwise connected iff A is an interval.
    4. Let ACR^n be pathwise connected, let f:A->R^n be continuous, then f(A) is PC.
    5. Every convex set is pathwise connected.

    That is all we have learn so far. Would that help?
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    Defintion:
    A set B not connected if there are open sets U and V such that:
    B interset U is not empty, B interset V is not empty.
    (B interset) U U (B interset V) equals to B.
    (B interset U) interset (B interset V) is empty.
    Did you look at the attached file?
    The above definition is exactly what was used in that proof.
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  6. #6
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    yeah, I just read it and it really does help. Thanks!
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