# Compact Sets in R^n .... .... D&K Theorem 1.8.4 .... ....

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In summary, the conversation discusses the proof of Theorem 1.8.4 from Duistermaat and Kolk's "Multidimensional Real Analysis I: Differentiation." It explains how to negate Apostol's definition of a bounded set to arrive at the statement that a sequence can be found satisfying certain conditions. It also discusses how this statement leads to the proof that the extraction of a convergent subsequence is impossible. The definition of compactness and its relevance to the conversation is also mentioned.
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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 an aspect of the proof of Theorem 1.8.4 ... ...

Duistermaat and Kolk"s Theorem 1.8.4 and its proof read as follows:https://www.physicsforums.com/attachments/7716
View attachment 7717
In the above proof we read the following:

Assume $$\displaystyle K$$ is not bounded, Then we can find a sequence $$\displaystyle ( x_k )_{ k \in \mathbb{N} }$$ satisfying $$\displaystyle x_k \in K$$ and $$\displaystyle \mid \mid x_k \mid \mid \ge k$$, for $$\displaystyle k \in \mathbb{N}$$. Obviously in this case the extraction of a convergent subsequence is impossible ... ... ... "
Question 1

Assuming Apostol's definition of a bounded set (D&K don't give one!) [see below for Apostol's definition] ... ... how do we logically and rigorously negate the definition of bounded set (since $$\displaystyle K$$ NOT bounded) and arrive at D&K's statement that then we can find a sequence $$\displaystyle ( x_k )_{ k \in \mathbb{N} }$$ satisfying $$\displaystyle x_k \in$$K and $$\displaystyle \mid \mid x_k \mid \mid \ge k$$, for $$\displaystyle k \in \mathbb{N}$$ ... ... ?Question 2

How do we formally and rigorously demonstrate that the statement " ... we can find a sequence $$\displaystyle ( x_k )_{ k \in \mathbb{N} }$$ satisfying $$\displaystyle x_k \in K$$ and $$\displaystyle \mid \mid x_k \mid \mid \ \ge k$$, for $$\displaystyle k \in \mathbb{N}$$ ... " leads to the statement ... " ... the extraction of a convergent subsequence is impossible ... ... ... " ... ... (note that although this seems plausible the rigorous demonstration that it is the case eludes me) ... ...Peter
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NOTE 1

Apostol's definition of a bounded set reads as follows:View attachment 7718NOTE 2D&K's definition of compactness and their development and comments regarding compactness may be helpful to MHB members reading the above post ... ... so I am providing the same ... as follows:
View attachment 7719

Peter said:
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 an aspect of the proof of Theorem 1.8.4 ... ...

Duistermaat and Kolk"s Theorem 1.8.4 and its proof read as follows:

In the above proof we read the following:

Assume $$\displaystyle K$$ is not bounded, Then we can find a sequence $$\displaystyle ( x_k )_{ k \in \mathbb{N} }$$ satisfying $$\displaystyle x_k \in K$$ and $$\displaystyle \mid \mid x_k \mid \mid \ge k$$, for $$\displaystyle k \in \mathbb{N}$$. Obviously in this case the extraction of a convergent subsequence is impossible ... ... ... "
Question 1

Assuming Apostol's definition of a bounded set (D&K don't give one!) [see below for Apostol's definition] ... ... how do we logically and rigorously negate the definition of bounded set (since $$\displaystyle K$$ NOT bounded) and arrive at D&K's statement that then we can find a sequence $$\displaystyle ( x_k )_{ k \in \mathbb{N} }$$ satisfying $$\displaystyle x_k \in$$K and $$\displaystyle \mid \mid x_k \mid \mid \ge k$$, for $$\displaystyle k \in \mathbb{N}$$ ... ... ?Question 2

How do we formally and rigorously demonstrate that the statement " ... we can find a sequence $$\displaystyle ( x_k )_{ k \in \mathbb{N} }$$ satisfying $$\displaystyle x_k \in K$$ and $$\displaystyle \mid \mid x_k \mid \mid \ \ge k$$, for $$\displaystyle k \in \mathbb{N}$$ ... " leads to the statement ... " ... the extraction of a convergent subsequence is impossible ... ... ... " ... ... (note that although this seems plausible the rigorous demonstration that it is the case eludes me) ... ...Peter
=========================================================================================

NOTE 1

Apostol's definition of a bounded set reads as follows:NOTE 2D&K's definition of compactness and their development and comments regarding compactness may be helpful to MHB members reading the above post ... ... so I am providing the same ... as follows:

A subset $K$ of $\mathbf R^n$ is bounded if there is $r>0$ such that $K\subseteq B(0, r)$, where $0$ is the origin. Therefore, if $K$ is not bounded, then for all natural numbers $k>0$, $K$ is not contained in $B(0, k)$, and thus there is $x_k\in K$ with $|x_k|>k$. This answers Question 1.

For Question 2, if we have a sequence $(x_k)_{k\in \mathbf N}$ in $\mathbf R^n$ with the property that $|x_k|>k$ for all $k$, then no subsequence of this seuqnce can converse. This is because of $(x_{k_r})_{r\in \mathbf N}$ is a convergent subsequence, then the sequence of naturals $(k_r)_{r\in \mathbf N}$ also converges (because the norm map $\|\cdot\|:\mathbf R^n\to \mathbf R$ is continuous). But any convergent sequence of reals is bounded and we have a contradiction.

caffeinemachine said:
A subset $K$ of $\mathbf R^n$ is bounded if there is $r>0$ such that $K\subseteq B(0, r)$, where $0$ is the origin. Therefore, if $K$ is not bounded, then for all natural numbers $k>0$, $K$ is not contained in $B(0, k)$, and thus there is $x_k\in K$ with $|x_k|>k$. This answers Question 1.

For Question 2, if we have a sequence $(x_k)_{k\in \mathbf N}$ in $\mathbf R^n$ with the property that $|x_k|>k$ for all $k$, then no subsequence of this seuqnce can converse. This is because of $(x_{k_r})_{r\in \mathbf N}$ is a convergent subsequence, then the sequence of naturals $(k_r)_{r\in \mathbf N}$ also converges (because the norm map $\|\cdot\|:\mathbf R^n\to \mathbf R$ is continuous). But any convergent sequence of reals is bounded and we have a contradiction.

Thanks for the help caffeinemachine ...

... really appreciate your help ...

Peter

## 1. What are compact sets in R^n?

Compact sets in R^n are subsets of the n-dimensional real numbers that are bounded and closed. This means that the set contains all of its limit points and has a finite size or extent.

## 2. What does the D&K Theorem 1.8.4 state?

The D&K Theorem 1.8.4, also known as the Heine-Borel Theorem, states that in R^n, a set is compact if and only if it is closed and bounded.

## 3. How is compactness related to convergence?

In general, compact sets can be thought of as "small" or "converging" sets, as they contain all their limit points. This means that any sequence of points in the set must have a limit point within the set itself, making it a useful concept when studying convergence.

## 4. Can compact sets exist in infinite-dimensional spaces?

Yes, compact sets can exist in infinite-dimensional spaces. In fact, the concept of compactness originated in infinite-dimensional spaces such as function spaces and is essential in many areas of mathematics, such as functional analysis and topology.

## 5. How are compact sets useful in mathematical analysis?

Compact sets are essential in mathematical analysis because they allow for the proof of important theorems, such as the Uniform Convergence Theorem and the Extreme Value Theorem. They also provide a way to study convergence and continuity in a general setting.

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