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Math Help - Continuity

  1. #1
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    Continuity

    Let g be defined on all of R. If A is a subset of R, define the set g^-1(A) by
    g^-1(A)={x in R : g(x) in A}.
    Show that g is continuous iff g^-1(O) is open whenever O contained in R is an open set.

    well g^-1(O) means g(O) is in A.
    Let g be continuous and O be an open subset of R.
    Then |x-c|<delta and |g(x)-g(c)|<epsilon

    Now I get stuck
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  2. #2
    Junior Member nimon's Avatar
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    "Then |x-c|<delta and |g(x)-g(c)|<epsilon."

    This just won't do. Write back with the proper definition of continuity and we'll carry on from there.
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  3. #3
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    A function f:A--->R is continuous at a point c in A if, for all epsilon epsilon>0, there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.
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  4. #4
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    I tried to get started on both directions:

    suppose g^-1(O) is open for each open subset O of R. Fix a real number x, and let epsilon be given. The neighborhood N centered at g(x) with radius epsilon is an open set, so the inverse image of N under g is an open set containing x. Thus there is a neighborhood V of x with radius delta such that V is a subset of g^-1(N)


    If g is continuous at x, then there exists a delta such that whenever y is less than delta apart from x, g(y) is in a neighborhood of g(x) with radius epsilon.
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  5. #5
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    Lets suppose that for each open set O\subset \mathbb{R} it is the case that f^{-1}(O) is also open.
    Given a point p\in\mathbb{R} and \varepsilon  > 0 then O=\left(f(p)- \varepsilon ,f(p)+ \varepsilon \right) is an open set.
    So Q=f^{-1}(O) is an open set that contains p.
    Hence, \left( {\exists \delta  > 0} \right) such that (p-\delta,p+\delta)\subset Q.
    But f(p-\delta,p+\delta)\subset O.
    Which by the \varepsilon/\delta definition f is continuous.
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