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

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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 ...

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
PeroK
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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
 
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PeroK
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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
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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