Open, Closed, and Half-Open Nests

Plato is right. OP sinks. Thanks. A consideration of what went wrong leads to the following:

**Open, Closed, and Half-Open Nests**

Given nests of intervals ai,bi = (), (], [), or [] and limlbi-ail=0 so that limai=limbi=a. When is there a point common to all intervals?

1)

limai=a but no ai=a

limbi=b but no bi=a

Then ai<a<bi, ai<a<=bi, ai<=a<bi, ai<=a<=bi and (), (], [), [] all have a common point.

2)

limai=a and ai=a, i>N

limbi=a but no bi=a

Is there an x st ai<x<bi or a<x<=bi i>N? NO. So nests (ai,bi) and (ai,bi] have no common point. But ai<=a<=bi and ai<=b<bi so [ai,bi], [ai,bi) have a common point.

3)

limai=a but no ai=a

limbi=a and bi=a, i>M

Then (ai,bi) and [ai,bi) have no common point, but [ai,bi] and (ai,bi] have a common point. Same reasoning as 2).

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Hartlw** A consideration of what went wrong leads to the following:

**Open, Closed, and Half-Open Nests**

Given nests of intervals ai,bi = (), (], [), or [] and limlbi-ail=0 so that limai=limbi=a. When is there a point common to all intervals?

The real difficulty with the OP is the word *nests*

There are cases where that concept is very general, see Kelly.

But you seem to restrict consideration to real numbers.

Saying that $\displaystyle \left\{ {\left( {a_n ,b_n } \right)} \right\}_n $ is a nested sequence of sets usually means that if $\displaystyle k<j$ then $\displaystyle [a_j,b_j]\subset (a_k,b_k)$.

If that is what *nest* means then the sequence $\displaystyle (a_n)$ is increasing, bounded above; the sequence $\displaystyle (b_n)$ is decreasing , bounded below. In both cases the sequences converge and the limit(s) belong to the common part of the nest.

QUESTION: Do you mean something different by nest?

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Plato** The real difficulty with the OP is the word *nests*

There are cases where that concept is very general, see Kelly.

But you seem to restrict consideration to real numbers.

Saying that $\displaystyle \left\{ {\left( {a_n ,b_n } \right)} \right\}_n $ is a nested sequence of sets usually means that if $\displaystyle k<j$ then $\displaystyle (a_j,b_j)\subset (a_k,b_k)$.

If that is what *nest* means then the sequence $\displaystyle (a_n)$ is increasing, bounded above; the sequence $\displaystyle (b_n)$ is decreasing , bounded below. In both cases the sequences converge and the limit(s) belong to the common part of the nest.

QUESTION: Do you mean something different by nest?

I assumed familiarity with the text-book def of nests of real numbers. By a nest ai,bi I mean ai+1>=ai and bi+1<=bi, and of course ai+1,bi+1 contained in ai,bi and limlbi-ail = 0 from which it follows that ai has a glb and bi has a lub and they are equal, but that's fluff and not really relevant, whats relevant is that they have the same limit. I was very careful to distinquish all the different possibilites of convergence of differenent types of intervals which I have never seen done in textbooks and I always wondered why the intervals had to be closed and what would happen if they were'nt. That question has now been answered, for the first time as far as I know, by my previous post.

The closest any text book comes to explaining why the intervals have to be closed in order to have a common point is to give a counter example. However, as I have shown, depending on situation, open and half open intervals can also have a point in common, especially in the fairly general case where ai<a and a<bi all i when they will always have a point in common.

I didn't take the time to describe exactly with math typography what is available in any text under "nested intervals" as it would take me forever to copy it from a text. As I said, I assumed famliarity with nested intervals, and besides, I'm lousy at TEX. I'm sure you can google the classical proof of intersection of __closed__ nested intervals.

edit: The crucial question is nests of real numbers. Nests in n-space then determine real nests of the individual coordinates.

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Hartlw** The closest any text book comes to explaining why the intervals have to be closed in order to have a common point is to give a counter example.

Why not post the counter-example.

It may help to see exactly what you are asking.

Frankly, I am confused as to your actual question.

POST SCRIPT.

If you consider $\displaystyle \left[ {1 - \frac{1}{n},1} \right)$ to fit your description, ie

$\displaystyle a_n=1 - \frac{1}{n}~\&~b_n=1$ then both sequences converge to one but the intersection is empty. The limit point is not in any of the sets. So again the exact description of a nest makes a difference.

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Plato** Why not post the counter-example.

It may help to see exactly what you are asking.

Frankly, I am confused as to your actual question.

I'm not at home, and I really don't feel like looking it up and copying it. And besides, a counter-example is irrelevant as I have carefully covered all the cases in post #6, which replaces the OP.

As for my actual question: why do std proofs that the intersection of closed intervals have to be closed never explain why other than to give a counter example?

OK, a counter example occurs to me. The nest (0,1/n) has no common point but [0,1/n] does. This case is covered in the complete theorem on nested intervals in post #6

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Hartlw** OK, a counter example occurs to me. The nest (0,1/n) has no common point but [0,1/n] does. This case is covered in the complete theorem on nested intervals in post #6

In the above $\displaystyle \left(0, \frac{1}{n}\right)$ the limit point $\displaystyle 0$ belongs to none of the sets. The intersection is empty.

On the other hand for $\displaystyle \left[0, \frac{1}{n}\right)$ the limit point $\displaystyle 0$ belongs to **ALL** of the sets. So the intersection is not empty. But the sets are not closed either.

So to insure an nonempty intersection the sets in the nest must be closed. You can prove that.

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Plato** In the above $\displaystyle \left(0, \frac{1}{n}\right)$ the limit point $\displaystyle 0$ belongs to none of the sets. The intersection is empty.

On the other hand for $\displaystyle \left[0, \frac{1}{n}\right)$ the limit point $\displaystyle 0$ belongs to **ALL** of the sets. So the intersection is not empty. But the sets are not closed either.

So to insure an nonempty intersection the sets in the nest must be closed. You can prove that.

That's what I just said in previous post ( I don't see the number): "OK, a counter example occurs to me. The nest (0,1/n) has no common point but [0,1/n] does." Why are you repeating it? And anyhow, the proof resides in post#6 which includes the general case.

Re: Open, Closed, and Half-Open Nests

Quote:

Originally Posted by

**Hartlw** The nest (0,1/n) has no common point but [0,1/n] does." Why are you repeating it? And anyhow, the proof resides in post#6 which includes the general case.

The reason is simple: I cannot read your notation.

**Please learn to use LateX** to avoid this kind of confusion in the future.

Re: Open, Closed, and Half-Open Nests

The standard theorem says that, using closed intervals, we can ensure a common point, belongs to all of the nested intervals.

And the counter example of (0,1/n] says, using non-closed nested intervals, we can't ensure there is a common point, belongs to all of them.

That's it. This explains why "closed" is required by the theorem.

I didn't see any problem here.