1. ## Heine-Borel Theorem

Hi,

I am trying to understand the first proof "A set S of real numbers is compact if and only if every open cover C of S can be reduced to a finite subcovering. " on this page: Theorem 5.2.6: Heine-Borel Theorem

But I don't understand towards the end when they say "However, aN+1 is an element of S, so that this subcovering can not cover S." How do we know aN+1 is an element of S?

2. That is an incredibly non-standard way of defining a compact set. Compact sets have the property that every open cover has a finite subcover. This is the best way of defining compact sets, because it is more generally applicable than defining compact sets as closed and bounded, as the web page you've linked to appears to do.

You know that $a_{N+1}\in S$ by construction. Look at the bullet point:

$\text{ for any }n>1\;\text{there exists }\mathbf{a_{n}\in S}\;\text{with...}$

3. "Compact", defined as "every open cover has a finite subcover", can be defined in any topological space. It can be shown that any compact set is closed. Even to define "bounded" requires a metric space rather than a general topological space but in a metric space, it can be proven that any compact set is bounded.

The other way, that all closed and bounded sets are "compact" (in this sense) requires the real number system or space derived from the real numbers because the proof requires the "completeness" property of the real numbers. That is why Ackbeet says that "closed and bounded" is "an incredibly non-standard way of defining a compact set".

4. hmmm okay, I guess I'm just trying to visualize how they justify that

for any n > 1 there exists an S with | s - an | < 1 / n because every neighborhood of s must contain elements from S.

I understand that every point in S has a neighbourhood in S, but why does this property allow us to make An sufficiently large?

(We defined a compact set in class like they did too... my prof said compact means closed an bounded, and so he wanted us to show how it was closed by contradiction, and so this link looks like a similiar method, but I'm just trying to understand it.)

5. why does this property allow us to make An sufficiently large?
They're not "sufficiently large". In fact, the $a_{n}$'s are getting closer and closer to $s$.

my prof said compact means closed an bounded
In the more typical "compact defined as open covers have finite subcovers" definition, compact always implies closed and bounded, but in some metric spaces, closed and bounded does not imply compact. Compact is a stronger condition in some metric spaces.

6. Originally Posted by Ackbeet
They're not "sufficiently large". In fact, the $a_{n}$'s are getting closer and closer to $s$.
Ohh okay, so for example, (s - A1000) < 1/1000 ... so the An is just any point getting closer and closer to s. So now, if C = { comp([s - 1/n, s + 1/n]), n > 0 } ... C should be finite and open and includes everything but s? So I don't understand why C couldn't include An+1. I must be picturing it wrong. Like, I picture S as something like

S: ---(-------)--- last bracket ")" would be where s is located.

And C: --------)(------------ where s i located at )( .

So why can't C include all points around s like S?

7. C should be finite and open and includes everything but s?
C is not a finite collection of sets, it is an infinite collection of sets. Every set in the collection C is open, because every set in the collection C is the complement of a closed set. For any point in S that is not s, it is in at least one set in the collection C, because

$\cup C=S\setminus\{s\}.$

C contains a set that includes $A_{N+1}.$ C contains at least one set that includes all of the $A_{n}$'s. However, because C is an open cover of S, and because you have assumed that every open cover of S has a finite subcover, that implies that $A_{N+1}$ is not in the finite subcover that you have assumed exists. That is the main point here. You have then found a point in S that is not in the supposedly finite subcover. Hence it's not really a finite subcover, because it doesn't contain every point of S.

Make sense?