Compact Subsets of R .... Sohrab, Proposition 4.1.1 (Lindelof) ....

[Math Amateur]

TL;DR

I am quite perplexed by Sohrab's proof of the Lindelof Covering Theorem ... any clear explanations of the strategy and tactics of the proof are very welcome ...

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 4: Topology of $\mathbb{R}$ and Continuity ... ...

I need help in order to fully understand the proof of Proposition 4.1.1...Proposition 4.1.1, some preliminary notes and its proof read as follows:

My questions are as follows:Question 1

In the above proof by Sohrab we read the following:

" ... ... Now the set $\{ \rho_x \ : \ x \in O \} \subset \mathbb{Q}$ is countable ... ... "

But ... it seems to me that since the $x$'s are uncountable that the number of $\rho_x$ is uncountable ... but that many (at times infinitely many ... ) have the same values since each is equal to a rational number and these are countable ...

... so in fact there are an uncountably infinite number of open balls $B_{ \rho_x } (x)$ ... there are just a countable number of different values for the radii of the open balls ...

Is my interpretation correct ... ?

Question 2

In the above proof by Sohrab we read the following:

" ... ... If for each $k \in \mathbb{N}$ we pick $\lambda_k \in B_{ \rho_k } (x_k) \subset O_{ \lambda_k }$, then we have a countable subcollection $\{ O_{ \lambda_k } \}_{ k \in \mathbb{N} } \subset \{ O_\lambda \}_{ \lambda \in \Lambda }$ which satisfies $O = \bigcup_{ k = 1 }^{ \infty } O_{ \lambda_k }$ ... ..."Can someone please explain/demonstrate clearly (preferably in some detail) how the process described actually results in a countable subcollection where $O = \bigcup_{ k = 1 }^{ \infty } O_{ \lambda_k }$ ...
In addition to answers to the two questions, any explanations/clarifications of the overall strategy and tactics of the proof would be very gratefully received ...
Peter

 

[andrewkirk]

For question 1 you are correct. The proof should have instead used the set

$$\{\rho_x\ :\ x\in\mathbf O\cap \mathbb Q\}$$

in both cases where that set is used.

However then it is no longer obvious that the union of $B_{\rho_x}(x)$ for $x\in \mathbf O\cap \mathbb Q$ covers $\mathbf O$. So I think the proof is fundamentally flawed. At a minimum, a few more steps will be needed to patch over that. We will use the density of the rationals, but we can't just say 'because of density' and wave our hand, as the author appears to have done.

 

[Math Amateur]

Thanks Andrew ...

Appreciate your help ...

Peter

 

[andrewkirk]

Actually the problem can be fixed fairly easily. After the first two sentences of the proof, we write:

Since the rational numbers are dense in $\mathbb R$, for each $x\in\mathbf O$ we can find $y_x,z_x\in\mathbb Q$ such that $x\in B_{y_x}(z_x)\subseteq B_{\varepsilon_x}(x)\subseteq O_{\lambda_x}\subseteq\mathbf O$.

Let $Y = \{y_x\ :\ x\in\mathbf O\}$ and $Z = \{z_x\ :\ x\in\mathbf O\}$.

Then $\bigcup_{y_x\in Y}\bigcup_{z_x\in Z} B_{y_x}(z_x)$ is a countable union of intervals that is equal to $\mathbf O$. Index the intervals in the union by the natural numbers as $I_1,I_2,...$.
Then for each $I_k$ we choose $\lambda_k$ such that $I_k\subseteq O_{\lambda_k}$.

Then $\mathbf O = \bigcup_{k\in\omega} I_k = \bigcup_{k\in\omega} O_{\lambda_k} \subseteq \mathbf O$,
whence $\bigcup_{k\in\omega} O_{\lambda_k} = \mathbf O$ as required.

 

[member 587159]

For question 1 you are correct. The proof should have instead used the set

$$\{\rho_x\ :\ x\in\mathbf O\cap \mathbb Q\}$$

in both cases where that set is used.

However then it is no longer obvious that the union of $B_{\rho_x}(x)$ for $x\in \mathbf O\cap \mathbb Q$ covers $\mathbf O$. So I think the proof is fundamentally flawed. At a minimum, a few more steps will be needed to patch over that. We will use the density of the rationals, but we can't just say 'because of density' and wave our hand, as the author appears to have done.

I disagree with this answer. The set $\{p_x\mid x\in \mathcal{O}\}$ is a subset of the rationals and therefore countable. It doesn't matter that it is indexed by an uncountable set!

 

[andrewkirk]

The set $\{p_x\mid x\in \mathcal{O}\}$ is a subset of the rationals and therefore countable.

Fair comment. I agree. The problem actually comes later in the proof, where they state without proof that
$$\bigcup_{k\in\omega} O_{\lambda_k} = \mathbf O$$
The containment $\subseteq$ is easy, but I suspect the reverse containment $\supseteq$ cannot go through with the definitions they have made. I'm happy to be corrected.

 

[member 587159]

Fair comment. I agree. The problem actually comes later in the proof, where they state without proof that
$$\bigcup_{k\in\omega} O_{\lambda_k} = \mathbf O$$
The containment $\subseteq$ is easy, but I suspect the reverse containment $\supseteq$ cannot go through with the definitions they have made. I'm happy to be corrected.

You are correct. The problem occurs later indeed. I think the proof can be fixed though halving some radii and playing with the triangle inequality. But I think your argument works as well.

 

[Math Amateur]

Actually the problem can be fixed fairly easily. After the first two sentences of the proof, we write:

Since the rational numbers are dense in $\mathbb R$, for each $x\in\mathbf O$ we can find $y_x,z_x\in\mathbb Q$ such that $x\in B_{y_x}(z_x)\subseteq B_{\varepsilon_x}(x)\subseteq O_{\lambda_x}\subseteq\mathbf O$.

Let $Y = \{y_x\ :\ x\in\mathbf O\}$ and $Z = \{z_x\ :\ x\in\mathbf O\}$.

Then $\bigcup_{y_x\in Y}\bigcup_{z_x\in Z} B_{y_x}(z_x)$ is a countable union of intervals that is equal to $\mathbf O$. Index the intervals in the union by the natural numbers as $I_1,I_2,...$.
Then for each $I_k$ we choose $\lambda_k$ such that $I_k\subseteq O_{\lambda_k}$.

Then $\mathbf O = \bigcup_{k\in\omega} I_k = \bigcup_{k\in\omega} O_{\lambda_k} \subseteq \mathbf O$,
whence $\bigcup_{k\in\omega} O_{\lambda_k} = \mathbf O$ as required.

Hi Andrew ...

The above seems a reasonable fix to me ... are you still standing by the validity of the above ...

Just a clarification ...

In the above you write:

" ... ... $\mathbf O = \bigcup_{k\in\omega} I_k$ ... ... "

Can you please explain why/how this is true ...

By the way ... what is $\omega$ ... ?

Thanks again to you and QED for your help ...

Peter

 

[andrewkirk]

$\omega$ is sometimes used to denote the natural numbers. On reflection, I don't know why I didn't just use the more common $\mathbb N$.

Why the above thing is true:

The first direction:
$\forall k:\ I_k\subseteq O_{\lambda_k}\subseteq \mathbf O$
so
$\bigcup_{k\in\mathbb N} I_k\subseteq \mathbf O$.

The reverse direction:

For each $x\in\mathbf O$ we have $x\in B_{y_x}(z_x)=I_k$ for some $k$
(because $I:\mathbb N\to \{B_y(z)\ :\ y\in\mathbb Q_+, z\in \mathbb Q\}$ such that $I(k)=I_k$ was chosen to be surjective, which we know is possible because $\mathbb Q_+\times \mathbb Q$ is countable).

So $x\in \bigcup_{j\in\mathbb N}I_j$. Since $x$ was arbitrary, we have $\mathbf O \subseteq \bigcup_{j\in\mathbb N}I_j$, as required.

The above uses the axiom of choice. It is possible that is not necessary, but constructing a proof that does not rely on AC would take more effort.