# Thread: Another Question on the Heine-Borel Theorem in R^n ... ... D&K Theorem 1.8.17 ... ...

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

2. 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 $\mathbb{R}^n \setminus K$.

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 $\mathbb{R}^n \setminus K$.