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

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In summary: Yes, I was talking about \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."
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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:

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

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"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.

HallsofIvy said:
"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

## 1. What is the Heine-Borel Theorem in R^n?

The Heine-Borel Theorem in R^n is a fundamental result in real analysis that states that a subset of R^n is compact if and only if it is closed and bounded. This theorem is named after mathematicians Eduard Heine and Émile Borel, who independently proved it in the late 19th century.

## 2. How is the Heine-Borel Theorem used in mathematics?

The Heine-Borel Theorem is used in various areas of mathematics, including real analysis, topology, and functional analysis. It is particularly useful in proving the existence of solutions to equations and in establishing the convergence of sequences and series.

## 3. What is the significance of the Heine-Borel Theorem in R^n?

The Heine-Borel Theorem is significant because it provides a necessary and sufficient condition for a subset of R^n to be compact. This allows mathematicians to easily identify and work with compact sets, which have many important properties and applications in mathematics.

## 4. How does Stromberg's Theorem 3.40 relate to the Heine-Borel Theorem?

Stromberg's Theorem 3.40 is a generalization of the Heine-Borel Theorem in R^n. It states that a subset of a metric space is compact if and only if it is sequentially compact, meaning that every sequence in the subset has a convergent subsequence. This theorem encompasses the Heine-Borel Theorem as a special case when the metric space is R^n.

## 5. Can the Heine-Borel Theorem be extended to other spaces?

Yes, the Heine-Borel Theorem has been extended to other spaces, such as metric spaces, topological spaces, and normed vector spaces. These extensions are known as the Heine-Borel Theorem for general spaces, and they provide necessary and sufficient conditions for a subset to be compact in these different types of spaces.

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