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Math Help - Continuous function on metric spaces.

  1. #1
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    Continuous function on metric spaces.

    Let X, Y be metric spaces and f: X->Y be a continous function on the indicated metric spaces. Prove the followings.

    (a) For each subset A of X, f(\overline {A}) \subset \overline{f(A)}.
    (b) For each subset B of Y, f^{-1}(int B) \subset int f^{-1}(B).

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    In my text, "int A" denotes the interior of A, which is the set of all interior points of A. A point x in A is an interior point of A, or A is neighborhood of x, provided that there is an open set O which contains x and is contained in A.
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    Let X, Y be metric spaces and f: X->Y be a continous function on the indicated metric spaces. Prove the followings.

    (a) For each subset A of X, f(\overline {A}) \subseteq \overline{f(A)}.
    We need a fact: a\in \bar A if and only if B_X(a,\epsilon) \cap A \not = \emptyset for any \epsilon > 0.
    Here, B_X(a,\epsilon) = \{x\in X |  d_X(x,a) < \epsilon \} i.e. the ball in X at a with radius \epsilon.

    Let y\in f(\bar A) this means y = f(a) for some a\in \bar A. Let \epsilon > 0.
    There is \delta > 0 such that if x\in B_X(a,\delta) \implies d_Y(f(x),f(a)) < \epsilon.
    However, B_X(a,\delta) \cap A \not = \emptyset and so pick some x\in B_X(a,\delta) \text{ and }x\in A.
    Therefore, f(x) \in f(A) and f(x) \in B_Y(f(a),\epsilon).
    Thus, f(A) \cap B_Y(y,\epsilon) \not = \emptyset for any \epsilon>0.
    This means by above fact that y\in \overline{f(A)}.
    Thus, f(\bar A)\subseteq \overline{f(A)}.
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