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Math Help - Open Set Proving

  1. #1
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    Red face Open Set Proving

    The question is as follows:

    Let f:R->R be continuous. Show that {x:f(x)>0} is an open subset of R.

    I have figured out how to prove {x:f(x)=0} is a closed subset of R, but I can not solve the question {x:f(x)>0} is an open one.

    Any help for that?
    Thanks.
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  2. #2
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    If f is continuous at x=a~~f(a)>0 in the definition of continuity use \varepsilon  = \frac{{f(a)}}{2}.

    It follows that f(x)>\frac{{f(a)}}{2} for all x near a.
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  3. #3
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    Thanks Plato, I am considering what you said.

    Quote Originally Posted by Plato View Post
    If f is continuous at x=a~~f(a)>0 in the definition of continuity use \varepsilon  = \frac{{f(a)}}{2}.

    It follows that f(x)>\frac{{f(a)}}{2} for all x near a.
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  4. #4
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    Thanks, and I get what you said about f(x) > f(a)/2 for all x near a, but why does it support {x:f(x)>0} is an open set?
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  5. #5
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    In order to show that a set is open, you must know the definition.

    One characterization of an open is that if a is in then every point near a is also in the set.
    Another is that open sets are union of balls.
    Another is that its complement is closed.
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  6. #6
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    I get it! Thanks very much. The proving part is tricky.
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