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Math Help - Right continuous

  1. #1
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    Right continuous

    Notation:
    (\mathbb{R},\tau_s) Standard topology on R
    (\mathbb{R}.\tau_l) Lower limit topology on R

    Prove f:\mathbb{R}\to\mathbb{R} is right continuous on \mathbb{R} iff. f is continuous on \mathbb{R}_l when considered as a function from \mathbb{R}_l\to\mathbb{R}.

    (\Rightarrow)
    Since f is right continuous on R, we know \lim_{x\to x^+_0}f(x)=f(x_0)
    f^{-1}((a_i,b_i))=\cup_{i\in I}[a_i,b_i)
    Where (a_i,b_i) is open in X. I am not sure about the saying the invers equals the union of half closed and open sets. I am sort of lost.
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    MHF Contributor Drexel28's Avatar
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    Re: Right continuous

    Quote Originally Posted by dwsmith View Post
    Notation:
    (\mathbb{R},\tau_s) Standard topology on R
    (\mathbb{R}.\tau_l) Lower limit topology on R

    Prove f:\mathbb{R}\to\mathbb{R} is right continuous on \mathbb{R} iff. f is continuous on \mathbb{R}_l when considered as a function from \mathbb{R}_l\to\mathbb{R}.

    (\Rightarrow)
    Since f is right continuous on R, we know \lim_{x\to x^+_0}f(x)=f(x_0)
    f^{-1}((a_i,b_i))=\cup_{i\in I}[a_i,b_i)
    Where (a_i,b_i) is open in X. I am not sure about the saying the invers equals the union of half closed and open sets. I am sort of lost.
    So, you've got the right idea. Right continuity on \mathbb{R} is equivalent to saying that \displaystyle f^{-1}(a,b)=\bigcup [c,d), but this right hand union is a union of sets open in \mathbb{R}_\ell...so.
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    Re: Right continuous

    Quote Originally Posted by Drexel28 View Post
    So, you've got the right idea. Right continuity on \mathbb{R} is equivalent to saying that \displaystyle f^{-1}(a,b)=\bigcup [c,d), but this right hand union is a union of sets open in \mathbb{R}_\ell...so.
    I don't know.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Right continuous

    Quote Originally Posted by dwsmith View Post
    I don't know.
    Yes you do. The right hand side of that last equality is a union of open sets in \mathbb{R}_\ell. What's true about a union of open sets?
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    Re: Right continuous

    Quote Originally Posted by Drexel28 View Post
    Yes you do. The right hand side of that last equality is a union of open sets in \mathbb{R}_\ell. What's true about a union of open sets?
    It just seems to easy. What about the other direction? I have no idea what to do for that one.
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    MHF Contributor Drexel28's Avatar
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    Re: Right continuous

    Quote Originally Posted by dwsmith View Post
    It just seems to easy. What about the other direction? I have no idea what to do for that one.
    Well, do you have an idea? You use the same formula. Just show that the preimage of an interval looking like that implies that the right limit is the actual value.
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  7. #7
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    Re: Right continuous

    Quote Originally Posted by Drexel28 View Post
    Well, do you have an idea? You use the same formula. Just show that the preimage of an interval looking like that implies that the right limit is the actual value.
    f^{-1}((a_i,b_i))=[a_i,b_i) which is open since f is continuous. By the well ordering principle, \bigcup [a_i,bi) has a least element say y.

    Therefore, \lim_{x\to y^+}f(x)=f(y)

    Does this work?
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