B Is the Explanation of Compactness in [0,1] and (0,1) Accurate?

[BWV]

Is this statement accurate? (Not trying a proof, just trying to understand the concept)

[0,1] is compact because any infinite collection of open subsets that covers the set has to include the boundry points therefore there is a finite collection of these sets that also covers

(0,1) is not compact because to cover the set you need an infinite number of subsets to obtain the limit condition on the open boundary - with a finite collection of open subsets you can always arbitrarily find a point outside of the union of your finite collection of open subsets and closer to a boundary

 

[mfb]

[0,1] is compact, but your argument does not show that there must be such a finite collection.

(0,1) is not compact because to cover the set you need an infinite number of subsets

No, there are finite covers of it. (0,1) itself is a trivial example.

You have to show that there are infinite covers that don't have a finite subcover, e.g. by giving an example.

 

[BWV]

Thanks, the definition is that every infinite cover contains a finite subcover so:

(0,1) is not compact because there exists infinite covers that relay on the limit condition on the open boundary - with a finite collection of open subsets you can always arbitrarily find a point outside of the union of your finite collection of open subsets and closer to a boundary

My thinking was that with the closed interval you did not have these infinite covers that relied on limits, not a proof in itself but maybe could be used in one?

 

[fresh_42]

My thinking was that with the closed interval you did not have these infinite covers that relied on limits, not a proof in itself but maybe could be used in one?

Only if you can make a mathematical statement out of it. "does not rely on" is far too vague to be used anywhere. You can cover $[0,1]$ by infinite many sets which don't allow a finite selection, e.g. $\bigcup_{n\in \mathbb{N}} \,(\frac{1}{n},1-\frac{1}{n}) \cup \{0\} \cup \{1\}$.

 

[BWV]

Only if you can make a mathematical statement out of it. "does not rely on" is far too vague to be used anywhere. You can cover $[0,1]$ by infinite many sets which don't allow a finite selection, e.g. $\bigcup_{n\in \mathbb{N}} \,(\frac{1}{n},1-\frac{1}{n}) \cup \{0\} \cup \{1\}$.

Ok, but if [0,1] is compact doesn’t that mean any infinite cover allows a finite selection?

 

[fresh_42]

Ok, but if [0,1] is compact doesn’t that mean any infinite cover allows a finite selection?

Yes, it does. The truth is hidden behind what a cover is. It is requested to be open sets, which wasn't the case in my example. Thus it is usually called an open cover and an open finite subcover. Finiteness alone isn't the point here.

 

[BWV]

Thanks,

Right, so it has to be covered by open sets, so the issue is that to cover a closed set with an group of open sets there have to be some sets that contain the boundary. A proof that [0,1] is compact would consist of demonstrating that any infinite collection of open sets that covers the interval contains a finite cover.

So
$\bigcup_{n\in \mathbb{N}} \,(\frac{1}{n} -1/2,1-\frac{1}{n}+1/2)$

Is an infinite cover of [0,1] which allows
n in {2,3} as a finite subcover

This would also cover (0,1) but that’s OK because compact means that all infinite covers allow a finite subcover and per above you can cover an open set with collections that require limits at the boundariesCorrect?

 

[Stephen Tashi]

Correct?

You are developing the correct intuition, but your focus on boundaries could be misleading.

As a subset of the real numbers, is the set $\mathbb{Z}$ consisting of the integers compact?

 

[BWV]

Before I googled it, I would have thought that the set of integers in R was an open set - I don’t get how the concept of boundary applies to Z. The beginning analysis book I am reading has limited the discussion to finite sunsets of R

 

[jbriggs444]

Before I googled it, I would have thought that the set of integers in R was an open set - I don’t get how the concept of boundary applies to Z. The beginning analysis book I am reading has limited the discussion to finite sunsets of R

When in doubt, look at the definition.

"A subset

of a topological space

is compact if for every open cover of

there exists a finite subcover of

"

Nothing about boundaries in there. So, can you think of an infinite open cover without a finite open subcover?

 

[BWV]

But is not every compact space bounded (Heine–Borel theorem)?
Googling, it looks like the set of integers in R is closed, but unbounded therefore not compact?

you can't have a finite number of open sets covering the infinite set of integers, correct?

 

[jbriggs444]

you can't have a finite number of open sets covering the infinite set of integers, correct?

Sure you can. The set of all real numbers is an open set.

[But it may not be a subcover of the infinite cover that I asked you to think of]

 

[member 587159]

Can I ask where your intuition about boundary points comes from? Does the definition of open covering say something about boundary points, or are you maybe using a certain theorem...?

 

[BWV]

Well its something I am trying to understand - what intuition I have is the result of trying to read a book on introductory analysis last night.

the Wikipedia entry on the Heine-Borel theorem states:

For a subset S of Euclidean space Rn, the following two statements are equivalent:

• S is closed and bounded
• S is compact, that is, every open cover of S has a finite subcover. This is the defining property of compact sets, called the Heine–Borel property.

 

[member 587159]

Well its something I am trying to understand - what intuition I have is the result of trying to read a book on introductory analysis last night.

the Wikipedia entry on the Heine-Borel theorem states:

For a subset S of Euclidean space Rn, the following two statements are equivalent:

• S is closed and bounded
• S is compact, that is, every open cover of S has a finite subcover. This is the defining property of compact sets, called the Heine–Borel property.

Yes, that's indeed where the intuition should come from. What book are you reading on analysis?

 

[BWV]

 

[mfb]

So
$\bigcup_{n\in \mathbb{N}} \,(\frac{1}{n} -1/2,1-\frac{1}{n}+1/2)$

Is an infinite cover of [0,1] which allows
n in {2,3} as a finite subcover

Okay, you checked a single example of an infinite cover. But you have to check all infinite covers to make sure they all have a finite subcover.

 

[BWV]

Okay, you checked a single example of an infinite cover. But you have to check all infinite covers to make sure they all have a finite subcover.

Well, one at a time. Will post the rest of them over the next few days

 

[mathwonk]

that is a great post.

 

[FactChecker]

The main skill to learn in a theoretical math class is to be very precise and disciplined with your definitions and proofs. So trying to get an informal intuition is the wrong thing to do till after you have a very clear understanding of the precise definitions and proofs. First, you need to learn to use the definitions, proofs, and theorems.

 

[member 587159]

The main skill to learn in a theoretical math class is to be very precise and disciplined with your definitions and proofs. So trying to get an informal intuition is the wrong thing to do till after you have a very clear understanding of the precise definitions and proofs. First, you need to learn to use the definitions, proofs, and theorems.

I don't know if I agree with this. Intuition is often very useful to memorize definitions, and I think it is useful to have some intuition before defining things formally (E.g. limit of a sequence, understanding what's going on makes you understand why the definition looks that way, and why the quantifiers are not in another order). But I agree that relying on intuition can be very misleading, especially, since this post is about analysis, in general metric spaces that look nothing alike the Euclidean space.

As an easy example, intuition could lead to the conjecture that the closure of open balls are closed balls, but this is false (if I remember correctly the discrete space gives a counterexample), while the intuition in the Euclidean space might tell you that this is true.

 

[FactChecker]

I don't know if I agree with this. Intuition is often very useful to memorize definitions, and I think it is useful to have some intuition before defining things formally (E.g. limit of a sequence, understanding what's going on makes you understand why the definition looks that way, and why the quantifiers are not in another order). But I agree that relying on intuition can be very misleading, especially, since this post is about analysis, in general metric spaces that look nothing alike the Euclidean space.

As an easy example, intuition could lead to the conjecture that the closure of open balls are closed balls, but this is false (if I remember correctly the discrete space gives a counterexample), while the intuition in the Euclidean space might tell you that this is true.

Ok. I'll buy that.

 

[WWGD]

I am sort of in-between both approaches. I used to be somewhat -fanatical about learning from the bottom-up, but I now think that there are too-many variables involved in the learning process to have a fixed strategy and I do bottom-up, top-down, sideways and any other combination that seems to work. EDIT: It comes down to input; just expose yourself to the material in as many forms as possible and take it from there, adjust given what you understand. Just expose yourself and tweak. EDIT: I just don't think one can really understand, model they way in which the brain/mind will tie things together. So , to me, repeated exposure is the best approach, let your subconscious guide you there.

 

[mathwonk]

Maybe you can improve your intuition by proving that a subset S of R is compact if and only if every continuous function f:S-->R is bounded. The definition of continuity plus the open cover version of compactness is useful for going one direction, i.e. for proving this boundedness property is implied by compactness. The other direction is easier to prove using the closed and bounded version.

I.e. try proving: heine borel implies all continuous functions are bounded implies closed and bounded.

Then the tricky part is proving all closed and bounded subsets of R have the heine borel property. The usual proof chooses a bounded closed interval [a,b] I containing the set S and, given an open cover of S, looks at the set of all t such that the subset S intersect [a,t] can be covered by a finite sub cover. Then look at the "least upper bound" of such t, and prove it equals b.

 

[WWGD]

Is this statement accurate? (Not trying a proof, just trying to understand the concept)

[0,1] is compact because any infinite collection of open subsets that covers the set has to include the boundry points therefore there is a finite collection of these sets that also covers

(0,1) is not compact because to cover the set you need an infinite number of subsets to obtain the limit condition on the open boundary - with a finite collection of open subsets you can always arbitrarily find a point outside of the union of your finite collection of open subsets and closer to a boundary

I don't know if you are interested in this perspective, for metric spaces, a space is compact ( to compare/contrast with other spaces such as (0,1) or [0,1] or any other type of subset) if every sequence has a convergent subsequence. A clear example of when this does not hold is for the Integers as a subset of the Real line. EDIT: If it interests you, you can try to prove some sort of equivalence of proofs using sequences with proofs using open covers. Still, it seems like a good idea to use open sets alone because the definition with open sets is a universal one; more general than the one with sequences.