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Thread: Inverse of an open set is open implies continuous

  1. #1
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    Inverse of an open set is open implies continuous

    A theorem in my book says:

    Suppose that $\displaystyle (X_1 , \rho \ , \ (X_2, \gamma ) $ are metric spaces. Define a function $\displaystyle f: X_1 \rightarrow X_2 $. Then f is continuous iff $\displaystyle f^{-1} (U) $ is open in $\displaystyle X_1$ for every open set $\displaystyle U \subset X_2 $.

    Proof in the book goes:

    Assume that for each open $\displaystyle U \subset X_2 $, $\displaystyle f^{-1}(U) $ is open in $\displaystyle X_1$.

    Let $\displaystyle x \in X_1 $, then $\displaystyle f(x) \in X_2 $.

    Then $\displaystyle \forall \epsilon > 0$, $\displaystyle f^{-1} (B(f(x), \epsilon )) $ is open in $\displaystyle X_1 $ since $\displaystyle B(f(x), \epsilon ) $ is open in $\displaystyle X_2$, therefore f is continuous.

    Q1: I understand why the set in the inverse is open, but how does that implies f is continuous? In other words, if I fill in the details, say:

    Let $\displaystyle x_0 \in X_1 $, then $\displaystyle \forall \epsilon > 0 $, pick $\displaystyle \delta > 0$, then whenever $\displaystyle \rho (x_0, x ) < \delta \ \ \ \ \ for \ x \in X_1 $, I need to get $\displaystyle \gamma ((f(x_0),f(x))) < \epsilon $ But how did the statement from above help me to get there?

    ---------------

    Conversely, suppose that f is continuous, let U be open set in $\displaystyle X_2$.

    $\displaystyle \forall y \in U $, there exists $\displaystyle \epsilon _y > 0 \ s.t. \ B( \epsilon _y , y ) \subset U $,
    and $\displaystyle \forall x \in f^{-1}( \{ y \} ) , \exists \delta _x > 0 $

    such that $\displaystyle B( \delta _x , x ) \subset f^{-1}(B( \epsilon _y , y ) ) \subset f^{-1} (U) $

    Implies that $\displaystyle f^{-1}(U) = \bigcup _{x \in f^{-1}(U) } B( \delta _x , x ) $ is open.

    Q2: I understand that the open ball around y is in U since it is an open set, why I'm a bit unsure about how to get $\displaystyle B( \delta _x , x) \subset f^{-1}(B( \epsilon _y , y )) $, I'm sure it is because f is continuous. I know that continuous function map an open set to an open set, but do I know for sure it is true for the inverse?

    Thank you!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    A theorem in my book says:

    Suppose that $\displaystyle (X_1 , \rho \ , \ (X_2, \gamma ) $ are metric spaces. Define a function $\displaystyle f: X_1 \rightarrow X_2 $. Then f is continuous iff $\displaystyle f^{-1} (U) $ is open in $\displaystyle X_1$ for every open set $\displaystyle U \subset X_2 $.

    Proof in the book goes:

    Assume that for each open $\displaystyle U \subset X_2 $, $\displaystyle f^{-1}(U) $ is open in $\displaystyle X_1$.

    Let $\displaystyle x \in X_1 $, then $\displaystyle f(x) \in X_2 $.

    Then $\displaystyle \forall \epsilon > 0$, $\displaystyle f^{-1} (B(f(x), \epsilon )) $ is open in $\displaystyle X_1 $ since $\displaystyle B(f(x), \epsilon ) $ is open in $\displaystyle X_2$, therefore f is continuous.

    Q1: I understand why the set in the inverse is open, but how does that implies f is continuous? In other words, if I fill in the details, say:

    Let $\displaystyle x_0 \in X_1 $, then $\displaystyle \forall \epsilon > 0 $, pick $\displaystyle \delta > 0$, then whenever $\displaystyle \rho (x_0, x ) < \delta \ \ \ \ \ for \ x \in X_1 $, I need to get $\displaystyle \gamma ((f(x_0),f(x))) < \epsilon $ But how did the statement from above help me to get there?

    ---------------

    Conversely, suppose that f is continuous, let U be open set in $\displaystyle X_2$.

    $\displaystyle \forall y \in U $, there exists $\displaystyle \epsilon _y > 0 \ s.t. \ B( \epsilon _y , y ) \subset U $,
    and $\displaystyle \forall x \in f^{-1}( \{ y \} ) , \exists \delta _x > 0 $

    such that $\displaystyle B( \delta _x , x ) \subset f^{-1}(B( \epsilon _y , y ) ) \subset f^{-1} (U) $

    Implies that $\displaystyle f^{-1}(U) = \bigcup _{x \in f^{-1}(U) } B( \delta _x , x ) $ is open.

    Q2: I understand that the open ball around y is in U since it is an open set, why I'm a bit unsure about how to get $\displaystyle B( \delta _x , x) \subset f^{-1}(B( \epsilon _y , y )) $, I'm sure it is because f is continuous. I know that continuous function map an open set to an open set, but do I know for sure it is true for the inverse?
    Then you know something that is not true! For example, f(x)= a, that is, the function that is maps all x to the fixed point a, is certainly continuous but f(U) for any set U, including open sets, is the singleton set {a} which is NOT open. I suspect you are simply remembering, incorrectly, the fact that the continuous inverse image of an open set is open.

    Thank you!
    Every ball, $\displaystyle B(\epsilon_y,y)$ is open, of course, so $\displaystyle f^{-1}(B(\epsilon_y, y))$ is open. That means that every point in the set, in particular x itself, is an "interior point" and so there exist neighborhood [itex]B(\delta, x)\subset f^{-1}(B(\epsilon_y, y)[/itex]. That is all you need.
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