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

In summary, the problem with the proof is that they use a set that is not countable and then try to use the density of the rationals to prove that it is.
  • #1
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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:
Sohrab - Proposition 4.1.1 ... .png
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
 
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  • #2
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.
 
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  • #3
Thanks Andrew ...

Appreciate your help ...

Peter
 
  • #4
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.
 
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  • #5
andrewkirk said:
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!
 
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  • #6
Math_QED said:
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.
 
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  • #7
andrewkirk said:
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.
 
  • #8
andrewkirk said:
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
 
  • #9
##\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.
 
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FAQ: Compact Subsets of R .... Sohrab, Proposition 4.1.1 (Lindelof) ....

1. What is a compact subset of R?

A compact subset of R is a subset of the real numbers that is both closed and bounded. This means that the subset contains all of its limit points and can be contained within a finite interval.

2. How is compactness defined in mathematics?

In mathematics, compactness is a property of a set that describes its ability to be covered by a finite number of smaller sets. A compact set is one that can be covered by a finite number of open sets, also known as an open cover.

3. What is the significance of Proposition 4.1.1 (Lindelof) in relation to compact subsets of R?

Proposition 4.1.1 (Lindelof) states that every open cover of a compact subset of R has a finite subcover. This means that any open cover of a compact subset can be reduced to a finite number of open sets that still cover the subset. This is a fundamental property of compact sets and is often used in proofs related to compactness.

4. How is compactness related to continuity?

In mathematics, continuity is a property of functions that describes their ability to preserve the structure of a set. Compactness is often used in the definition of continuity, as a function is continuous if and only if the preimage of any open set is a compact set.

5. What are some real-world applications of compact subsets of R?

Compact subsets of R have many real-world applications, particularly in fields such as physics, engineering, and economics. For example, compact sets are used to model physical systems with finite energy, to describe the behavior of materials under stress, and to analyze the convergence of economic models. They are also used in optimization problems to find the minimum or maximum value of a function within a given set.

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