Is the Completeness Axiom Equivalent to the Heine Borel Theory?

[autobot.d]

Hello,

I am trying to prove that The Completeness Axiom is equivalent to the Heine Borel Theory (open finite cover).

I was wondering when going from Completeness Axiom to Heine Borel, is ok to assume that the set is closed?
I know the heine borel assumes that it is closed, but Completeness does not. I was just wondering if I had to prove that Completeness also asserted that the set was closed.

Thanks for the help.

 

[Landau]

What is your "completeness axiom"? That every set of real numbers bounded from above (below) has an upper (lower) bound? What set do you mean when you say "that Completeness also asserted THE set was closed"?

Also, Heine-Borel states equivalence between compact and (closed, bounded) subsets of R^n. It does not "assume" some set is closed.

 

[autobot.d]

That be the completeness axiom I am talking about.

Not sure what you mean by not assuming set is closed.

"Let F be a closed and bounded set of real numbers. Then every open cover of F has a finite subcover."

Thanks for the help!

 

[Landau]

Let me put it this way: yes, you have to assume F is closed and bounded, and then in your proof use the completeness axiom somewhere. The completeness axiom does not say anything about a set being closed.

 

[autobot.d]

Ahhh.

So I can only use the fact that my set is bounded, if I have you correct.
That is what I figured. Thank you again for your help.

 

[Landau]

you have to assume f is closed and bounded

so i can only use the fact that my set is bounded, if i have you correct.

I don't see how you get that from my answer.

 

[autobot.d]

The completeness axiom does not say anything about a set being closed.

I thought you meant that proving the Completeness Axiom using Heine Borel Theorem that I cannot use that the set is closed since the Completeness Axiom doesn't call for it. Sorry for the confusion.