Compact Subsets of R .... Sohrab, Proposition 4.1.8 .... ....

In summary: That means that there is some distance between ##\xi## and the rest of the cover, which can be used to get a contradiction.In summary, the conversation is about a question regarding the proof of Proposition 4.1.8 in Houshang H. Sohrab's book "Basic Real Analysis" (Second Edition). The question asks for help in understanding how to demonstrate rigorously that the open cover in the proof has no finite subcover. The conversation also includes a discussion about the importance of actively engaging in learning Real Analysis and finding an idea or strategy for a proof before translating it into rigorous work.
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Question in respect of the proof that compact subsets of R are necessarily closed and bounded ... issue of an open cover with no finite subcover ...
I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 4: Topology of R and Continuity ... ...

I need help in order to fully understand the proof of Proposition 4.1.8 ...Proposition 4.1.8 and its proof read as follows:
Sohrab - Proposition 4.1.8 ... .png

In the above proof by Sohrab we read the following:

" ... ... If, to get a contradiction, we assume that ##\xi \notin K## is a limit point of ##K##, then the open cover ##\{ ( - \infty, \xi - 1/n ) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} }## has no finite subcover ... ... "
My question is as follows:

How would we demonstrate rigorously that the open cover ##\{ ( - \infty, \xi - 1/n) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} }## has no finite subcover ... ...?
Help will be appreciated ...

Peter
=======================================================================================It may help readers of the above post to have access to Sohrab's definition of a limit point ... so I am providing the relevant text ... as follows ...
Sohrab - Definition 2.2.11 ... Limit Point  ... .png

Hope that helps ...

Peter
 
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Math Amateur said:
Summary: Question in respect of the proof that compact subsets of R are necessarily closed and bounded ... issue of an open cover with no finite subcover ...
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In the above proof by Sohrab we read the following:

" ... ... If, to get a contradiction, we assume that ##\xi \notin K## is a limit point of ##K##, then the open cover ##\{ ( - \infty, \xi - 1/n ) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} }## has no finite subcover ... ... "
My question is as follows:

How would we demonstrate rigorously that the open cover ##\{ ( - \infty, \xi - 1/n) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} }## has no finite subcover ... ...?

This should be in homework and you should be presenting your attempt at the solution. One thing you definitely can't learn vicariously is Real Analysis. You have to get stuck into these proofs yourself.
 
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Thanks for the reply PeroK ...

You write:

" ... ... One thing you definitely can't learn vicariously is Real Analysis. ... ... "

I agree ... but I couldn't get a meaningful start on the issue ...

I am still reflecting on the matter and reading other books' similar theorems and proofs to get some ideas ...

Peter
 
  • #4
Math Amateur said:
Thanks for the reply PeroK ...

You write:

" ... ... One thing you definitely can't learn vicariously is Real Analysis. ... ... "

I agree ... but I couldn't get a meaningful start on the issue ...

I am still reflecting on the matter and reading other books' similar theorems and proofs to get some ideas ...

Peter

You need an idea or strategy for a proof like this. Then , you need to translate the idea into rigorous work.

My first thought was that any finite subcover must have a greatest ##n##, which means the union cannot get arbitrarily close to ##\xi##.
 
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Related to Compact Subsets of R .... Sohrab, Proposition 4.1.8 .... ....

1. What is a compact subset of R?

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

2. How is compactness defined in Sohrab, Proposition 4.1.8?

In Sohrab, Proposition 4.1.8, compactness is defined as a property of a subset of R that satisfies the Heine-Borel theorem. This means that the subset is closed and bounded, and therefore contains all of its limit points.

3. What is the significance of Proposition 4.1.8 in Sohrab?

Proposition 4.1.8 in Sohrab is significant because it provides a rigorous definition of compactness in R, which is an important concept in analysis and topology. It also serves as a foundation for further proofs and theorems related to compact subsets of R.

4. How is compactness related to continuity?

Compactness and continuity are closely related concepts. In fact, a function is continuous on a compact subset of R if and only if it is uniformly continuous. This means that compactness is a necessary condition for continuity in certain cases.

5. Can you provide an example of a compact subset of R?

One example of a compact subset of R is the closed interval [0,1]. This subset is bounded by 0 and 1, and it contains all of its limit points. Therefore, it satisfies the definition of compactness in Sohrab, Proposition 4.1.8.

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