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  1. MHB Master
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    #1
    I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

    I am focused on Chapter 1: Continuity ... ...

    I need help with another aspect of the proof of Theorem 1.8.17 ... ...

    Duistermaat and Kolk's Theorem 1.8.17 and its proof (including the preceding relevant definition) read as follows:







    In the last line of the above proof we read the following:

    " ... ... and so $ \displaystyle y \in \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} \subset \mathbb{R}^n\text{\\} K$. ... ... "


    Presumably $ \displaystyle y \in \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} $

    ... because $ \displaystyle \mid \mid y - y \mid \mid = \mid \mid 0 \mid \mid \lt \frac{1}{ j_0 }$ ...

    Is that right?

    BUT ...

    How/why does $ \displaystyle y \in \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} \subset \mathbb{R}^n\text{\\}K$ mean that $ \displaystyle \mathbb{R}^n\text{\\} K$ is open?


    Help will be much appreciated ...

    Peter
    Last edited by Peter; February 11th, 2018 at 02:30.

  2. MHB Craftsman
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    #2
    Quote Originally Posted by Peter View Post
    I
    In the last line of the above proof we read the following:

    " ... ... and so $ \displaystyle y \in \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} \subset \mathbb{R}^n\text{\\} K$. ... ... "


    Presumably $ \displaystyle y \in \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} $

    ... because $ \displaystyle \mid \mid y - y \mid \mid = \mid \mid 0 \mid \mid \lt \frac{1}{ j_0 }$ ...

    Is that right?
    Yes, you are right. The set $\{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} $ is just the open ball of radius $\frac{1}{j_0}$ centered at $y$, so certainly $y$ is in it.

    Quote Originally Posted by Peter View Post
    BUT ...

    How/why does $ \displaystyle y \in \{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} \subset \mathbb{R}^n\text{\\}K$ mean that $ \displaystyle \mathbb{R}^n\text{\\} K$ is open?
    We started with an arbitrary point $y \in \mathbb{R}^n \setminus K$ and have now shown that there exists an open ball around $y$ that is entirely contained in $\mathbb{R}^n \setminus K$.

    So, each point of $\mathbb{R}^n \setminus K$ is an interior point of $\mathbb{R}^n \setminus K$. This means that $\mathbb{R}^n \setminus K$ is open.

    Another way to say the same thing: $\mathbb{R}^n \setminus K$ can be written as the union of open balls (choose an open ball for each such point $y$ as above and take the union of all such balls). Since any union (finite, infinite, doesn't matter) of open sets is open, this proves openness of $\mathbb{R}^n \setminus K$.

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by Krylov View Post
    Yes, you are right. The set $\{ x \in \mathbb{R}^n \ \mid \ \mid \mid x - y \mid \mid \lt \frac{1}{ j_0 } \} $ is just the open ball of radius $\frac{1}{j_0}$ centered at $y$, so certainly $y$ is in it.



    We started with an arbitrary point $y \in \mathbb{R}^n \setminus K$ and have now shown that there exists an open ball around $y$ that is entirely contained in $\mathbb{R}^n \setminus K$.

    So, each point of $\mathbb{R}^n \setminus K$ is an interior point of $\mathbb{R}^n \setminus K$. This means that $\mathbb{R}^n \setminus K$ is open.

    Another way to say the same thing: $\mathbb{R}^n \setminus K$ can be written as the union of open balls (choose an open ball for each such point $y$ as above and take the union of all such balls). Since any union (finite, infinite, doesn't matter) of open sets is open, this proves openness of $\mathbb{R}^n \setminus K$.

    Thanks Krylov... your post was extremely helpful...

    Appreciate your help and guidance...

    Peter

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