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Math Help - Continous function in topologies

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

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

    defined by

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

    defined by

    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 T_u and T_{\text{cof}}? The usual and cofinite topologies? And I assume R=\mathbb{R}.

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

    I also agree that it can't be continuous. For, if f:\left(\mathbb{R},T_u\right)\to\left(\mathbb{R},T  _{\text{cof}}\right) were continuous then f:\left(\mathbb{R},T_u\right)\to\left(\{-1,1\},T_s\right) where 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, \left(\{-1,1\},T_s\right)\approx D where D is the two point discrete space.


    Thus, if f were continuous there would be a map \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

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

    we will face a contradiction

    suppose D= {1,2}

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

    and \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 \bf{not} connected contradiction
    Exactement.
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