How do I show that the real numbers are not compact?

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Homework Help Overview

The discussion revolves around proving that the real numbers are not compact, specifically using the definition of compactness related to open covers and finite subcovers. Participants are exploring the implications of this definition and the characteristics of the real numbers in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find an open cover of the real numbers that cannot be reduced to a finite subcover. There are questions about what such an open cover might look like and whether specific examples would suffice.

Discussion Status

Several participants are actively engaging with the problem, suggesting potential open covers and clarifying definitions. There is a mix of ideas being explored, with some participants questioning the correctness of proposed sets and others refining their thoughts on how to construct an appropriate open cover.

Contextual Notes

Participants are working under the constraint of using the definition of compactness directly, which influences their approach to the problem. There is also a note of caution against posting complete solutions prematurely.

fred123
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A trivial, yet difficult question. How would one prove that the real numbers are not compact, only using the definition of being compact? In other words, what happens if we reduce an open cover of R to a finite cover of R?

I let V be a collection of open subset that cover R
Then I make the assumption that that this open cover can be reduced to a finite subcover.
(Clearly this is not possible) I am struggling to see/show what happens to R when I make this assumption. Should I simply find a counterexample, if so, what could it look like?

I have proven several, similar problems, but this one is so general that its tricky.
 
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You have to check two properties:
Closed?
Bounded?
What do you think?
 
But the question states that I should do this proof directly from the definition of being compact. Ie, if an open cover can be reduced to a finite subcover.
 
Compact: every open cover has a finite subcover.
So what is the definition of non-compact? Now show the reals satisfy this definition. If you can negate logical statements then this problem is not at all tricky.
 
So, find one open cover that cannot be reduced to a finite subcover?
But wouldn't that imply that I have to state what this open cover looks like? Should I maybe let R be an open cover of itself?
 
That open cover has exactly one set in it, R. One is a finite number.

Find an open cover of infinitely many sets that does not have a finite subcover. Post what you're thinking, i.e. how you might make such a cover.

Please, no one post a solution right away.
 
I think I sorted some stuff out, and {n-1,n+1 n in Z} I believe must work!? Thanks for the support
 
Well, I would write it (n-1,n+1) for n contained in Z. {n-1, n+1} means the set containing exactly two members, n-1 and n+1. That's not open. (I don't know what "n in Z" inside the braces could mean!)
 
You might try and have a closer look at the family of sets [itex]S=\{(-n,n),n\in\mathbb{N}\}[/itex]
 
  • #10
Fred's works too (and was essentially my first thought), assuming he meant { (n-1, n+1) | n in Z }.
 
  • #11
Agreed. I was unsure of what he meant with the type of parentheses he used.
 

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