# Continuous function on metric spaces.

• January 13th 2009, 02:08 AM
aliceinwonderland
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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• January 13th 2009, 12:24 PM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
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, $f(\bar A)\subseteq \overline{f(A)}$.