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Thread: Continous function in topologies

  1. #1
    MHF Contributor Amer's Avatar
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    Continous function in topologies

    $\displaystyle f: (R,T_u) \rightarrow (R,T_{cof}) $

    defined by

    $\displaystyle f(x)=\left\{ \begin {array}{cc} -1 ,& \mbox x\in Q \\ 1 , &\mbox x\in Q^c \end {array}\right.$


    is this function continuous, and is this function closed

    I found it closed and not continuous

    am I right ?

    Thanks in advance
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    $\displaystyle f: (R,T_u) \rightarrow (R,T_{cof}) $

    defined by

    $\displaystyle f(x)=\left\{ \begin {array}{cc} -1 ,& \mbox x\in Q \\ 1 , &\mbox x\in Q^c \end {array}\right.$


    is this function continuous, and is this function closed

    I found it closed and not continuous

    am I right ?

    Thanks in advance
    What are $\displaystyle T_u$ and $\displaystyle T_{\text{cof}}$? The usual and cofinite topologies? And I assume $\displaystyle R=\mathbb{R}$.

    I agree that it's closed. If $\displaystyle U\subseteq\mathbb{R}$ then $\displaystyle f(U)\subseteq\{-1,1\}$ and so $\displaystyle \mathbb{R}-f(U)\supseteq \mathbb{R}-\{1,-1\}$ is open. The image of any subset of $\displaystyle \mathbb{R}$ is closed.

    I also agree that it can't be continuous. For, if $\displaystyle f:\left(\mathbb{R},T_u\right)\to\left(\mathbb{R},T _{\text{cof}}\right)$ were continuous then $\displaystyle f:\left(\mathbb{R},T_u\right)\to\left(\{-1,1\},T_s\right)$ where $\displaystyle T_s$ is the subspace topology would be continuous. But, every subspace of a cofinite space has the cofinite topology, and a finite cofinite space is discrete. And so, $\displaystyle \left(\{-1,1\},T_s\right)\approx D$ where $\displaystyle D$ is the two point discrete space.


    Thus, if $\displaystyle f$ were continuous there would be a map $\displaystyle \varphi:\left(\mathbb{R},T_u\right)\to D$ which is surjective and continuous. What's the problem with that?
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  3. #3
    MHF Contributor Amer's Avatar
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    Yeah it is the usual and cofinite topologies and the set is the real numbers
    but I do not know how to write R with two lines

    about your question

    $\displaystyle \varphi : (\mathbb{R} ,T_u) \rightarrow D $

    we will face a contradiction

    suppose D= {1,2}

    $\displaystyle \varphi ^{-1} (1)$ and $\displaystyle \varphi ^{-1} (2) $ open sets in $\displaystyle \mathbb{R} $

    and $\displaystyle \varphi ^{-1} (1) \cup \varphi ^{-1} (2) = \mathbb{R} $

    that means real numbers with usual topology is connected contradiction
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    that means real numbers with usual topology is $\displaystyle \bf{not}$ connected contradiction
    Exactement.
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