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Math Help - continuous problem

  1. #1
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    Unhappy continuous problem

    could anyone help me with this problem please?
    Prove that the definition of continuous function f: R to R is equivalent to the property that for any open set A in R, the inverse-image of A is open.

    So I know the definition of a continuous function is that if for every x sub zero in R,and for every epsilon greater than 0, there exists delta such that the absolute value of f(x)-f(x_0) < epsilon if the absolute value of x-x_0 < delta.

    I don't know what kind of open set would have the inverse image is also open.
    I don't know how to approach this problem
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    could anyone help me with this problem please?
    Prove that the definition of continuous function f: R to R is equivalent to the property that for any open set A in R, the inverse-image of A is open.

    So I know the definition of a continuous function is that if for every x sub zero in R,and for every epsilon greater than 0, there exists delta such that the absolute value of f(x)-f(x_0) < epsilon if the absolute value of x-x_0 < delta.

    I don't know what kind of open set would have the inverse image is also open.
    I don't know how to approach this problem
    Say f is a continous function on \mathbb{R}. Let S be an open set we want to show f^{-1}(S) is also open. Note, if S=\{ \} then the proof is complete, so it is safe to say S is non-empty. Now f^{-1}(S) is open if and only if every point is an interior point, so we will show every point is an interior point. Let x_0\in f^{-1}(S) then f(x_0) \in S and so there exists a \epsilon > 0 so that all y satisfing |y-f(x_0)|<\epsilon lie in S. But continuity there is a \delta >0 so that |x-x_0|< \delta \implies |f(x)-f(x_0)|< \epsilon. That means by chosing \delta the set of points satisfing |x-x_0|<\delta all lie in f^{-1}(S). Q.E.D.
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  3. #3
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    Now say that f is a function show that the inverse-image of every open set is open. We want to show it is continous. Let x_0 \in \mathbb{R}. For \epsilon > 0 consider the set |y - f(x_0)|<\epsilon (meaning all y that satisfy this inequality). This set, say S, is clearly open. So its inverse image f^{-1}(S) is too open. Now x_0 \in f^{-1}(S) trivially. And so by hypothesis there is a \delta such that |x-x_0|<\delta \in f^{-1}(S) (meaning all x that satisfy this inequality). Thus, |x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon. Thus f is continous at any point x_0. Q.E.D.
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