Heine-Borel Theorem - Proof - Rudin Theorem 2.41

[Math Amateur]

I am reading Walter Rudin's book, Principles of Mathematical Analysis.

Currently I am studying Chapter 2:"Basic Topology".

Although I can basically follow it, I am concerned that I do not fully understand the proof of Theorem 2.41 (Heine-Borel Theorem).

Rudin, Theorem 2.41 reads as follows:View attachment 3795
View attachment 3796

In the above proof we read:

" ... It remains to be shown that (c) implies (a).

If $$E$$ is not bounded, then $$E$$ contains points $$x_n$$ with

$$| x_n | \gt n$$ $$ \ \ \ \ \ $$ $$(n = 1,2,3, ... )$$.

The set $$S$$ consisting of these points $$x_n$$ is infinite and clearly has no limit point in $$R^k$$, hence has none in $$E$$. ... ... "

I cannot see how Rudin concludes that the set $$S$$ "clearly" has no limit point in $$R^k$$ ... ...

Can someone explain exactly why this is the case ... what is the formal and rigorous argument?

PeterNOTE: I apologise to MHB members for the fact that a Mac Taskbar appears in the image above ... ... I have no idea how that happened!

 

[Math Amateur]

I am reading Walter Rudin's book, Principles of Mathematical Analysis.

Currently I am studying Chapter 2:"Basic Topology".

Although I can basically follow it, I am concerned that I do not fully understand the proof of Theorem 2.41 (Heine-Borel Theorem).

Rudin, Theorem 2.41 reads as follows:In the above proof we read:

" ... It remains to be shown that (c) implies (a).

If $$E$$ is not bounded, then $$E$$ contains points $$x_n$$ with

$$| x_n | \gt n$$ $$ \ \ \ \ \ $$ $$(n = 1,2,3, ... )$$.

The set $$S$$ consisting of these points $$x_n$$ is infinite and clearly has no limit point in $$R^k$$, hence has none in $$E$$. ... ... "

I cannot see how Rudin concludes that the set $$S$$ "clearly" has no limit point in $$R^k$$ ... ...
Can someone explain exactly why this is the case ... what is the formal and rigorous argument?

PeterNOTE: I apologise to MHB members for the fact that a Mac Taskbar appears in the image above ... ... I have no idea how that happened!

I have just reflected on the question in my above post ... I suspect that the answer is obvious ... my thoughts follow ... ...
We have $$S = \{ x_n \}$$ ...

Now, if some $$x_n$$ is a limit point of $$S$$, then every neighbourhood of $$x_n$$ must contain a point $$y \ne x_n$$ such that $$y \in S$$.

But the neighbourhood $$N_{1/2}(x_n)$$ consisting of points $$y \in S$$ such that $$d(x_n, y) \lt 1/2$$ contains no points except $$x_n$$ ... ...

So ... ... no $$x_n \in S$$ is a limit point of $$S$$ in $$R^k$$ ...
Can someone critique my analysis above and either confirm that the analysis is correct and/or point out any errors or shortcomings ...

Peter

 

[Euge]

You haven't proven that $S$ contains no limit point in $\Bbb R^k$. Even so, how do you know that $N_{1/2}(x_n)$ contains no point other than $x_n$? In your argument, you never used the fact that $|x_n| > n$ for all $n \in \Bbb Z^+$.

Now, if $S := \{x_n : n\in \Bbb Z^+\}$, then $S$ contains no limit point in $\Bbb R^k$ because $(x_n)$ has no convergent subsequence. Indeed, if $S$ had a limit point, then that point would be a limit of a subsequence $(x_{n_k})$ of $x_n$, which implies $(x_{n_k})$ is convergent. However, since $|x_n| > n$ for all $n$, for every subsequence $(x_{m_j})$ of $(x_n)$, $|x_{n_j}| > n_j \ge j$ for all $j$. This implies that every subsequence of $(x_{n_j})$ is unbounded, and consequently $x_n$ has no convergent subsequence. Therefore, $S$ has no limit point in $\Bbb R^k$.

 

[Math Amateur]

You haven't proven that $S$ contains no limit point in $\Bbb R^k$. Even so, how do you know that $N_{1/2}(x_n)$ contains no point other than $x_n$? In your argument, you never used the fact that $|x_n| > n$ for all $n \in \Bbb Z^+$.

Now, if $S := \{x_n : n\in \Bbb Z^+\}$, then $S$ contains no limit point in $\Bbb R^k$ because $(x_n)$ has no convergent subsequence. Indeed, if $S$ had a limit point, then that point would be a limit of a subsequence $(x_{n_k})$ of $x_n$, which implies $(x_{n_k})$ is convergent. However, since $|x_n| > n$ for all $n$, for every subsequence $(x_{m_j})$ of $(x_n)$, $|x_{n_j}| > n_j \ge j$ for all $j$. This implies that every subsequence of $(x_{n_j})$ is unbounded, and consequently $x_n$ has no convergent subsequence. Therefore, $S$ has no limit point in $\Bbb R^k$.

Thanks for your help, Euge.

By way of explanation (for some inexplicable reason! (Doh) ), I read Rudin's proof as saying that $$|x_n| = 1$$ and from there convinced myself that this meant that for each n, the open ball $$N_{1/2}(x_n)$$ would contain only $$x_n$$ ... ... and thus $$\{ x_n \}$$ would contain no limit points ... ... silly!

I am currently searching analysis texts to find theorems involving limit points and subsequences ... looks like you are saying there is a theorem that says that if a sequence has a limit point then it must be the limit of a subsequence of the sequence ...

Well, definitely sounds true, intuitively anyway ... just cannot find such a result in Rudin, Apostol or Pugh ...

But anyway, can follow your logic ...

Thanks again for your help,

PeterNOTE: I think Rudin must have had in mind a different/alternative argument from one involving subsequences since this proof is in Ch. 2: Basic Topology ... while the text deals with sequences and subsequences in Ch. 3: Numerical Sequences and Series ...

 

[Euge]

The "theorem" that you think I'm using is based off of the method of induction. Using the definition of a limit point of a sequence, you (inductively) extract a subsequence. Sequence extraction is something you have done several times when dealing with Noetherian rings and modules. With more practice, you will get a firmer grasp of the concept.

 

[Math Amateur]

The "theorem" that you think I'm using is based off of the method of induction. Using the definition of a limit point of a sequence, you (inductively) extract a subsequence. Sequence extraction is something you have done several times when dealing with Noetherian rings and modules. With more practice, you will get a firmer grasp of the concept.

Thanks Euge ... always appreciate your guidance and help ...

Peter