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Math Help - interior/ closure

  1. #1
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    interior/ closure

    Set A={f element C([0,1]): f(x)>0 for all x in [0,1]}
    Find the interior and closure of A in C([0,1])
    Explain

    Please help
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  2. #2
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    i am looking at the definitions for interior and kind of confused because it says that in order for S to be interior, Int(S) is an open set of the metric space -X

    looking at my metric space, C([0,1]) and my S, A, ...... it looks as thought that can't be true because C is a closed set

    grrrrrr
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by derek walcott View Post
    i am looking at the definitions for interior and kind of confused because it says that in order for S to be interior, Int(S) is an open set of the metric space -X

    looking at my metric space, C([0,1]) and my S, A, ...... it looks as thought that can't be true because C is a closed set

    grrrrrr
    A={f element C([0,1])

    when they write elements C([0,1]) that's means the elements of A is the continuous functions defined at [0,1]
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  4. #4
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    so is.... C([0,1]) a line segment on a closed interval then?

    and A is any point in C?

    i can't visualize it
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  5. #5
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    Int(A) = C((0,1))
    and closure of A = C([0,1])

    close... correct .... not close at all .....?
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  6. #6
    MHF Contributor Amer's Avatar
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    what is the whole question I think there is something missing,
    what is the base for the space we have.

    I solved it but I'm not sure ...

    if we say A is the subset of the space that contains all continuous functions at [0,1]
    and the functions of A is continuous and has a positive range

    A is open in that space since all elements of A is continuous functions at [0,1]

    so Int(A) =A

    Closure of A = A if A is closed in the space

    A is closed that means the complement of A is open

    the complement of A is the continuous functions defined at [0,1]
    with range equal 0 and negative range

    as you can see the complement is open since it contains continuous functions defined at [0,1]

    so Closure(A) = A since A is closed to

    ...
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  7. #7
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    professor never said anything about the base of the space
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  8. #8
    MHF Contributor Amer's Avatar
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    ok what is the metric defined at the space ??
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  9. #9
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    Letís get the notation straight. Here is the usual understanding for this material.
    \mathcal{C}([0,1]) is the set of continuous functions on [0,1].
    The metric is for \{f,g\}\subset\mathcal{C}([0,1]) then d(f,g) = {\max }_{x \in [0,1]}\left| {f(x) - g(x)} \right|.

    P.S.
    If f\in \mathcal{C}([0,1])\cap A the because of the properties of continuous functions we know that
    f has a positive minimum on [0,1]. Lets call it f(t).
    Now define \varepsilon  = \frac{{f(t)}}{2} > 0.
    Suppose h\in\mathcal{B}(f; \varepsilon )=\{g:d(f,g)< \varepsilon \}
    Can you show that h\in A?
    Last edited by Plato; April 25th 2010 at 03:37 PM.
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