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

[Math Amateur]

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 $$K$$ is not bounded, Then we can find a sequence $$( x_k )_{ k \in \mathbb{N} }$$ satisfying $$x_k \in K$$ and $$\mid \mid x_k \mid \mid \ge k$$, for $$k \in \mathbb{N}$$. 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 $$K$$ NOT bounded) and arrive at D&K's statement that then we can find a sequence $$( x_k )_{ k \in \mathbb{N} }$$ satisfying $$x_k \in $$K and $$\mid \mid x_k \mid \mid \ge k$$, for $$k \in \mathbb{N}$$ ... ... ?Question 2

How do we formally and rigorously demonstrate that the statement " ... we can find a sequence $$( x_k )_{ k \in \mathbb{N} }$$ satisfying $$x_k \in K$$ and $$\mid \mid x_k \mid \mid \ \ge k$$, for $$k \in \mathbb{N}$$ ... " 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

 

[caffeinemachine]

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 $$K$$ is not bounded, Then we can find a sequence $$( x_k )_{ k \in \mathbb{N} }$$ satisfying $$x_k \in K$$ and $$\mid \mid x_k \mid \mid \ge k$$, for $$k \in \mathbb{N}$$. 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 $$K$$ NOT bounded) and arrive at D&K's statement that then we can find a sequence $$( x_k )_{ k \in \mathbb{N} }$$ satisfying $$x_k \in $$K and $$\mid \mid x_k \mid \mid \ge k$$, for $$k \in \mathbb{N}$$ ... ... ?Question 2

How do we formally and rigorously demonstrate that the statement " ... we can find a sequence $$( x_k )_{ k \in \mathbb{N} }$$ satisfying $$x_k \in K$$ and $$\mid \mid x_k \mid \mid \ \ge k$$, for $$k \in \mathbb{N}$$ ... " 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.

 

[Math Amateur]

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