Heine-Borel Theorem in R^n .... Stromberg, Theorem 3.40 .... ....

[Math Amateur]

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:




View attachment 9141




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

 

[HallsofIvy]

"Because S is compact it must have a finite subcover."

No, that is not correct. In fact, it doesn't make sense. To talk about a "subcover" you must first have specified a "cover". Are you talking about the one given in the argument, $B_k(0)$, the collection of open balls with center at 0 and radius k= 1, 2, 3, ...?

Notice that these balls are each contained in the next so that any point in the union of a finite number of them is in the largest, the largest "k". Since every point is S is some finite distance from 0 every point of S is in one of those so it is an open cover for S.

Since S is compact there exist a finite

subcover of that cover. There are only a finite number of such "$B_k(0)$" so there is a largest "k". S is covered by that single set so the distance between any two points in S is less than 2k.​

​

 

[Math Amateur]

"Because S is compact it must have a finite subcover."

No, that is not correct. In fact, it doesn't make sense. To talk about a "subcover" you must first have specified a "cover". Are you talking about the one given in the argument, $B_k(0)$, the collection of open balls with center at 0 and radius k= 1, 2, 3, ...?

Notice that these balls are each contained in the next so that any point in the union of a finite number of them is in the largest, the largest "k". Since every point is S is some finite distance from 0 every point of S is in one of those so it is an open cover for S.

Since S is compact there exist a finite

subcover of that cover. There are only a finite number of such "$B_k(0)$" so there is a largest "k". S is covered by that single set so the distance between any two points in S is less than 2k.​

​

Thanks for the help, HallsofIvy ... ...

You write:

" ... ... No, that is not correct. In fact, it doesn't make sense. To talk about a "subcover" you must first have specified a "cover". Are you talking about the one given in the argument ... "


Yes, I was talking about \(\displaystyle \mathscr{U} = \{ B_k (0) \ : \ k \in \mathbb{N} \}\) as given by Stromberg ... but I did not say so explicitly ... so strictly speaking ... yes ... what I said was not correct ... indeed reading literally ... it didn't make sense ... as you say ...

Thanks again ...

Peter