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Thread: 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, $\displaystyle f(\overline {A}) \subset \overline{f(A)}.$
    (b) For each subset B of Y, $\displaystyle f^{-1}(int B) \subset int f^{-1}(B).$

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

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