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Math Help - Continuous mapping - closed, open, compact, bounded

  1. #1
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    Continuous mapping - closed, open, compact, bounded

    (I'm sorry to post in one thread, but I didn't want to cluster up the forums with multiple threads)

    The following are 8 possible outcomes of continuous mappings and its image and preimages. I have several questions regarding notations as well.
    ================================================== =================
    Let  A \subseteq \mathbb{R}^n and  F:A \rightarrow \mathbb{R} be continuous.

    Images:
    1. If A is closed, is F(A) closed?
    False. Let  A = \mathbb{R} and define  f(x) = tan^{-1} x . A is closed in R but  f(A) = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) is open and not closed.

    2. If A is open, is F(A) open?
    I believe this is true but I am not sure.

    3. If A is compact, is F(A) compact?
    True. Let  u_{k} be a sequence in F(A). Let  u_{k} = F(v_{k}) . Since A is sequentially compact, then there is a subsequence v_{k_j} \rightarrow v \in A. Thus,  {u_{k_j}} = {F(v_{k_j})}

    4. If A is bounded, is F(A) bounded?
    False. Let  A = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) and  f(x) = tanx. Since  A = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) is bounded, but  f:tan \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) \rightarrow R is unbounded.

    **If suppose it was f:R^n \rightarrow R^m is bounded, then would  f(A)= \mathbb{R}^m be bounded?


    Preimages:
    5. If A is closed, is F^{-1}(A) closed?
    I believe this is true. Since F^{-1}(A^{c}) = (f^{-1}(A))^c . So, if A is closed, then  A^{c} is open. If f is continuous, then the preimage of every closed set in R is closed in A. By definition.

    6. If A is open, is F^{-1}(A) open?
    True. If f is continuous, then the preimage of every open set in R is open in A by definition.

    7. If A is compact, is F^{-1}(A) compact?
    False, let A = [0, 1] which is compact because it is closed and bounded. Then let f(x) = 1/2, the the image  F(A) = \mathbb{R} which is not compact.

    8. If A is bounded, is F^{-1}(A) bounded?
    I do not know. I believe this is false, but I can not think of a counterexample.

    Regarding notations, what happens if instead of  f:A \rightarrow \mathbb{R} is continuous, it was f: \mathbb{R}^n \rightarrow \mathbb{R}^m is continuous. Would the results change?

    Thank you for reading. Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    (
    Images:
    1. If A is closed, is F(A) closed?
    False. Let  A = \mathbb{R} and define  f(x) = tan^{-1} x . A is closed in R but  f(A) = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) is open and not closed.
    This is false.

    2. If A is open, is F(A) open?
    I believe this is true but I am not sure.
    No, consider the constant function.

    3. If A is compact, is F(A) compact?
    True.
    Yes, this is true. This is a well-known result.

    4. If A is bounded, is F(A) bounded?
    False. Let  A = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) and  f(x) = tanx. Since  A = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) is bounded, but  f:tan \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) \rightarrow R is unbounded.
    False again.

    5. If A is closed, is F^{-1}(A) closed?
    I believe this is true. Since F^{-1}(A^{c}) = (f^{-1}(A))^c . So, if A is closed, then  A^{c} is open. If f is continuous, then the preimage of every closed set in R is closed in A. By definition.
    With pre-images this is true.

    6. If A is open, is F^{-1}(A) open?
    True. If f is continuous, then the preimage of every open set in R is open in A by definition.
    Again true.

    7. If A is compact, is F^{-1}(A) compact?
    False, let A = [0, 1] which is compact because it is closed and bounded. Then let f(x) = 1/2, the the image  F(A) = \mathbb{R} which is not compact.
    This is false since the next one is false.

    8. If A is bounded, is F^{-1}(A) bounded?
    I do not know. I believe this is false, but I can not think of a counterexample.
    Consider f:\mathbb{R}\to \mathbb{R} defined by f(x) = 0. Then \{ 0 \} is bounded but its preimage is unbounded.
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