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- Jun 22, 2012

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I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 3.40 on page 104 ... ...

Theorem 3.40 and its proof read as follows:

In the above proof by Stromberg we read the following:

" ... ... To see that \(\displaystyle S\) is bounded, consider \(\displaystyle \mathscr{U} = \{ B_k (0) \ : \ k \in \mathbb{N} \}\). Plainly \(\displaystyle \mathscr{U}\) is an open cover of \(\displaystyle \mathbb{R}^n\) and, in particular of \(\displaystyle S\). Thus there is a \(\displaystyle k_0 \in \mathbb{N}\) such that \(\displaystyle S \subset B_{ k_0 } (0)\); whence \(\displaystyle \text{diam}S \leq 2 k_0 \lt \infty\). ... ... "

I am troubled by the apparent (to me) lack of a rigorous argument to demonstrate that \(\displaystyle S\) is bounded ...

It appears to me that \(\displaystyle \mathbb{R}^n\) is "infinitely large", that is unbounded ... and hence \(\displaystyle S\) can be "infinitely large" and hence unbounded ... for example if \(\displaystyle S\) is the \(\displaystyle x_1\) axis of \(\displaystyle \mathbb{R}^n\) then \(\displaystyle S \subset \mathbb{R}^n\) and \(\displaystyle S\) is unbounded ...

Can someone please demonstrate rigorously that \(\displaystyle S\) is bounded ...

*** EDIT ***

Just a thought after reflection ... even though open covers of \mathbb{R}^n and S may be "infinite" (in dimension) because S is compact it must have a finite subcover ... and hence S is bounded ... is that the correct argument?

***************************************************************************************************

Help will be appreciated ...

Peter

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 3.40 on page 104 ... ...

Theorem 3.40 and its proof read as follows:

In the above proof by Stromberg we read the following:

" ... ... To see that \(\displaystyle S\) is bounded, consider \(\displaystyle \mathscr{U} = \{ B_k (0) \ : \ k \in \mathbb{N} \}\). Plainly \(\displaystyle \mathscr{U}\) is an open cover of \(\displaystyle \mathbb{R}^n\) and, in particular of \(\displaystyle S\). Thus there is a \(\displaystyle k_0 \in \mathbb{N}\) such that \(\displaystyle S \subset B_{ k_0 } (0)\); whence \(\displaystyle \text{diam}S \leq 2 k_0 \lt \infty\). ... ... "

I am troubled by the apparent (to me) lack of a rigorous argument to demonstrate that \(\displaystyle S\) is bounded ...

It appears to me that \(\displaystyle \mathbb{R}^n\) is "infinitely large", that is unbounded ... and hence \(\displaystyle S\) can be "infinitely large" and hence unbounded ... for example if \(\displaystyle S\) is the \(\displaystyle x_1\) axis of \(\displaystyle \mathbb{R}^n\) then \(\displaystyle S \subset \mathbb{R}^n\) and \(\displaystyle S\) is unbounded ...

Can someone please demonstrate rigorously that \(\displaystyle S\) is bounded ...

*** EDIT ***

Just a thought after reflection ... even though open covers of \mathbb{R}^n and S may be "infinite" (in dimension) because S is compact it must have a finite subcover ... and hence S is bounded ... is that the correct argument?

***************************************************************************************************

Help will be appreciated ...

Peter

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