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  1. MHB Master
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    #1
    I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

    I am focused on Chapter 5: Continuous Functions ...

    I need help in fully understanding an aspect of the proof of Theorem 5.3.2 ...


    Theorem 5.3.2 and its proof ... ... reads as follows:





    In the above text from Bartle and Sherbert we read the following:

    " Since $ \displaystyle I$ is closed and the elements of $ \displaystyle X'$ belong to $ \displaystyle I$, it follows from Theorem 3.2.6 that $ \displaystyle x \in I$. Then $ \displaystyle f$ is continuous at $ \displaystyle x$ ... ... "


    Can someone please explain exactly why/how we can conclude that $ \displaystyle f$ is continuous at $ \displaystyle x$ ... ?


    Peter

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    #2
    Quote Originally Posted by Peter View Post
    In the above text from Bartle and Sherbert we read the following:

    " Since $ \displaystyle I$ is closed and the elements of $ \displaystyle X'$ belong to $ \displaystyle I$, it follows from Theorem 3.2.6 that $ \displaystyle x \in I$. Then $ \displaystyle f$ is continuous at $ \displaystyle x$ ... ... "


    Can someone please explain exactly why/how we can conclude that $ \displaystyle f$ is continuous at $ \displaystyle x$ ... ?
    The theorem contains the hypothesis that $f$ is continuous on $I$. By definition, that means that $f$ is continuous at each point of $I$. The proof has already shown that $x\in I$. So it follows that $f$ is continuous at $x$.

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by Opalg View Post
    The theorem contains the hypothesis that $f$ is continuous on $I$. By definition, that means that $f$ is continuous at each point of $I$. The proof has already shown that $x\in I$. So it follows that $f$ is continuous at $x$.

    Oh ... careless of me not to notice that ...!

    Thanks Opalg ...

    Peter

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