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

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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0rthodontist
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.

matt grime
Homework Helper
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?

matt grime
Homework Helper
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

HallsofIvy
Homework Helper
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!)

arildno
Homework Helper
Gold Member
Dearly Missed
You might try and have a closer look at the family of sets $S=\{(-n,n),n\in\mathbb{N}\}$

Hurkyl
Staff Emeritus
Gold Member
Fred's works too (and was essentially my first thought), assuming he meant { (n-1, n+1) | n in Z }.

arildno