
MHB Master
#1
February 11th, 2018,
02:18
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.

February 11th, 2018 02:18
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MHB Craftsman
#2
February 11th, 2018,
08:01
Originally Posted by
Peter
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.
Originally Posted by
Peter
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 $\mathbb{R}^n \setminus K$.

MHB Master
#3
February 11th, 2018,
09:41
Thread Author
Originally Posted by
Krylov
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