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Math Help - Is the Interior of the space of continuous functions empty?

  1. #1
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    Is the Interior of the space of continuous functions empty?

    Let C be the space of continuous functions from [a,b] to R (real numbers) with the metric:

    d(f,g) = sup lf-gl

    Is the interior of C a non-empty set?


    I say it is an empty set, but I'm not sure about some of my steps. Let me know what you all think:

    Outline of Proof

    Suppose the interior is not empty and contains f.

    Then, for some e>0, the open ball around f of radius e is contained in C.

    But, because f is continuous, then if we break [a,b] up into small enough intervals such that on each interval [Xi, X(i+1)], there is a number Ci whereby:

    If x is in [Xi, X(i+1)) -->lf(x) - Cil < e

    Define a function g as:

    g(x) = Ci if x is in [Xi, X(i+1))

    then d(f,g) < e

    so g is in the ball of radius e around f

    so g is in C, but g is not continuous...contradiction.

    ...thoughts?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by southprkfan1 View Post
    Let C be the space of continuous functions from [a,b] to R (real numbers) with the metric:

    d(f,g) = sup lf-gl

    Is the interior of C a non-empty set?


    I say it is an empty set, but I'm not sure about some of my steps. Let me know what you all think:

    Outline of Proof

    Suppose the interior is not empty and contains f.

    Then, for some e>0, the open ball around f of radius e is contained in C.

    But, because f is continuous, then if we break [a,b] up into small enough intervals such that on each interval [Xi, X(i+1)], there is a number Ci whereby:

    If x is in [Xi, X(i+1)) -->lf(x) - Cil < e

    Define a function g as:

    g(x) = Ci if x is in [Xi, X(i+1))

    then d(f,g) < e

    so g is in the ball of radius e around f

    so g is in C, but g is not continuous...contradiction.

    ...thoughts?
    Clearly since \left(\mathcal{C}\left[a,b\right]\right)^{\circ}\subseteq\mathcal{C}\left[a,b\right] we must merely show that no point of \mathcal{C}[a,b] is an interior point.

    So, let f\in\mathcal{C}[a,b] and let \varepsilon>0 be given. Define, g(x)=\begin{cases} f(x) & \mbox{if} \quad x\ne\frac{a+b}{2} \\ f(x)+\frac{\varepsilon}{2} & \mbox{if} \quad x=\frac{a+b}{2}\end{cases}. Since \frac{a+b}{2}\in\left([a,b]\right)^{\circ} we know that it is continuous if and only if \lim_{x\to\frac{a+b}{2}}g(x)=g\left(\frac{a+b}{2}\  right), but \lim_{x\to\frac{a+b}{2}}g(x)=\lim_{x\to\frac{a+b}{  2}}f(x)=f\left(\frac{a+b}{2}\right)\ne g\left(\frac{a+b}{2}\right) and so g\notin\mathcal{C}\left[a,b\right]. But, g-f=\begin{cases}0 & \mbox{if} \quad x\ne \frac{a+b}{2} \\ \frac{\varepsilon}{2} & \mbox{if} \quad x=\frac{a+b}{2}\end{cases} and so \left\|g-f\right\|_{\infty}=\frac{\varepsilon}{2} and so g\in B_{\varepsilon}(f). The conclusion follows.

    Oh, I guess I tacitly assumed that a\ne b. Some authors, like the one of the book I'm reading now, allow a=b in which case [a,b]=\{a\} but in that case this is obvious.
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  3. #3
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    Thanks,

    So basically you agree with my method, its just that the way I got the non-continuous function g was way too complicated.
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  4. #4
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    Quote Originally Posted by southprkfan1 View Post
    Let C be the space of continuous functions from [a,b] to R (real numbers) with the metric:

    d(f,g) = sup lf-gl

    Is the interior of C a non-empty set?


    I say it is an empty set, but I'm not sure about some of my steps. Let me know what you all think:

    Outline of Proof

    Suppose the interior is not empty and contains f.

    Then, for some e>0, the open ball around f of radius e is contained in C.

    But, because f is continuous, then if we break [a,b] up into small enough intervals such that on each interval [Xi, X(i+1)], there is a number Ci whereby:

    If x is in [Xi, X(i+1)) -->lf(x) - Cil < e

    Define a function g as:

    g(x) = Ci if x is in [Xi, X(i+1))

    then d(f,g) < e

    so g is in the ball of radius e around f

    so g is in C, but g is not continuous...contradiction.

    ...thoughts?
    As stated this makes no sense: A topological space X is always open and so it is its own interior. Did you mean as a subspace of some larger space (ie. bounded functions on [a,b] with sup norm)?

    Also notice that both solutions are meaningless, since the functions constructed are not in the space you gave (thus the need for a larger one to fit in those).
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jose27 View Post
    As stated this makes no sense: A topological space X is always open and so it is its own interior. Did you mean as a subspace of some larger space (ie. bounded functions on [a,b] with sup norm)?

    Also notice that both solutions are meaningless, since the functions constructed are not in the space you gave (thus the need for a larger one to fit in those).
    I interpreted as being a subspace of bounded functions on [a,b]
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    I interpreted as being a subspace of bounded functions on [a,b]
    That's exactly right, I should have included that in the question. The actual question assumes we know this fact.
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  7. #7
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    I have posted a question very similar to this one that I would appreciate any help on:

    http://www.mathhelpforum.com/math-he...c-0-empty.html
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