# Heine-Borel Theorem & D&K Theorem 1.8.17 | Peter's Question

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In summary, the conversation discusses Duistermaat and Kolk's Theorem 1.8.17 and its proof, which involves showing that a set $\mathbb{R}^n\text{\\}K$ is open. The proof utilizes the concept of an open ball and shows that every point in $\mathbb{R}^n\text{\\}K$ is an interior point, proving its openness. The conversation also includes a question about why this proof shows the openness of $\mathbb{R}^n\text{\\}K$. The expert summarizer explains that this is because $\mathbb{R}^n\text{\\}K$ can be written as a union of
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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:View attachment 7765
View attachment 7766
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:
Peter said:
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.

Peter said:
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 $\mathbb{R}^n \setminus K$.

Krylov said:
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 $\mathbb{R}^n \setminus K$.

Peter

## 1. What is the Heine-Borel Theorem?

The Heine-Borel Theorem is a fundamental theorem in real analysis that states that a subset of Euclidean space is compact if and only if it is closed and bounded.

## 2. Who developed the Heine-Borel Theorem?

The Heine-Borel Theorem is named after German mathematicians Eduard Heine and Felix Borel, who independently proved the theorem in the late 19th century.

## 3. What is the significance of the Heine-Borel Theorem?

The Heine-Borel Theorem is significant because it provides a characterization of compact sets in Euclidean space, which is essential for many areas of mathematics such as analysis, topology, and geometry.

## 4. What is D&K Theorem 1.8.17?

D&K Theorem 1.8.17, also known as the Bolzano-Weierstrass Theorem, is a fundamental theorem in real analysis that states that every bounded sequence in Euclidean space has a convergent subsequence.

## 5. Who developed D&K Theorem 1.8.17?

D&K Theorem 1.8.17 is named after mathematicians Bernard Bolzano and Karl Weierstrass, who both contributed to the development of the theorem in the 19th century.

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