Can the intersection of nested closed intervals be empty?

[kingwinner]

Homework Statement

Nested interval theorem:
Suppose [a_n,b_n] is a nested sequence of closed intervals, i.e. [a_n+1,b_n+1] is contained in [a_n,b_n] for all n≥1. Then the intersection of all these intervals is nonempty.

Proof:
a_n is an increasing sequence bounded above by b₁ (or any b_n), so sup a_n = a exists and a≤b_m for all m.
So "a" is a lower bound of b_m
Thus let b = inf b_m, then we have a≤b (since any lower bound is ≤ its greatest lower bound).
Then [a,b] is contained in [a_n,b_n] for all n since a_n≤a and b≤b_n for all n (because a is an upper bound of a_n and b is a lower bound of b_n)
Therefore, the intersection is nonempty.
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1) In this theorem, it says that "...the seqeunce of CLOSED intervals...". Is it enough to have closed intervals, or must the endpoints also be FINITE numbers? Can any of the endpoints be -∞ or +∞? (reacall that intervals like [a,∞) are classified as closed interval)

2) In the proof, I am confused and I really don't understand why "a≤b_m for all m". Can somebody please explain this?

Homework Equations

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The Attempt at a Solution

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Thank you! :)

 

[iomtt6076]

1) In this theorem, it says that "...the seqeunce of CLOSED intervals...". Is it enough to have closed intervals, or must the endpoints also be FINITE numbers? Can any of the endpoints be -∞ or +∞? (reacall that intervals like [a,∞) are classified as closed interval)

$$\bigcap_{n=1}^\infty [n,\infty)=\emptyset$$

 

[iomtt6076]

2) In the proof, I am confused and I really don't understand why "a≤b_m for all m". Can somebody please explain this?

b_m is an upper bound for the sequence (a_n). By definition of the supremum, sup a_n is less than or equal to b_m

 

[kingwinner]

$$\bigcap_{n=1}^\infty [n,\infty)=\emptyset$$

But [n,∞) satisifies ALL assumptions as stated in the theorem (it is a nested sequence of closed intervals). So is the theorem wrong? What should be the correct statement of the theorem?

 

[iomtt6076]

But [n,∞) satisifies ALL assumptions as stated in the theorem (it is a nested sequence of closed intervals). So is the theorem wrong? What should be the correct statement of the theorem?

I think it should be "... a sequence of closed, bounded, non-empty intervals in $$\mathbb{R}$$..."

 

[kingwinner]

b_m is an upper bound for the sequence (a_n). By definition of the supremum, sup a_n is less than or equal to b_m

I agree that for each m, b_m is an upper bound for the sequence (a_n).
But I still don't understand why "sup a_n≤b_m for all m."
By definition of upper bound, sup a_n≥a_n for all n, but this idea seems to be unconnected with the above and I don't see how sup a_n can possibly be related to b_m. They are not even talking about the same sequence.

Can someone please explain this in more detail?

Thanks!

 

[iomtt6076]

I agree that for each m, b_m is an upper bound for the sequence (a_n).
But I still don't understand why "sup a_n≤b_m for all m."
By definition of upper bound, sup a_n≥a_n for all n, but this idea seems to be unconnected with the above and I don't see how sup a_n can possibly be related to b_m. They are not even talking about the same sequence.

Can someone please explain this in more detail?

Thanks!

Definition:
A number s is the supremum for a sequence if :
1) s is an upper bound for the sequence
2) if p is any upper bound for the sequence, s is less than or equal to p

so we are talking about 2) here. b_m is an upper bound for the sequence; thus by definition sup a_n<=b_m. It doesn't matter whether or not b_m is related to a sequence. All that matters is whether it is an upper bound. If this isn't good enough, then what about this: sup a_n is the least upper bound. suppose for contradiction b_m<sup a_n. remember that b_m is an upper bound for a_n. but then sup a_n wouldn't be the least upper bound because there is another upper bound, b_m, that is less than it. so we must have b_m>=sup a_n

 

[kingwinner]

Definition:
A number s is the supremum for a sequence if :
1) s is an upper bound for the sequence
2) if p is any upper bound for the sequence, s is less than or equal to p

so we are talking about 2) here. b_m is an upper bound for the sequence; thus by definition sup a_n<=b_m. It doesn't matter whether or not b_m is related to a sequence. All that matters is whether it is an upper bound.


If this isn't good enough, then what about this: sup a_n is the least upper bound. suppose for contradiction b_m<sup a_n. remember that b_m is an upper bound for a_n. but then sup a_n wouldn't be the least upper bound because there is another upper bound, b_m, that is less than it. so we must have b_m>=sup a_n

Thanks, so it just follows from the fact that the least upper bound of (a_n) is less than or equal to any upper bound of (a_n). Got it!:)

One more question...

How can we PROVE that b_m is an upper bound for the sequence (a_n) for ALL m=1,2,3,...?

When I draw a picture, I can see that this is the case, but I am not sure how to write it out and PROVE rigorously.
We know that
a1≤a2≤a3≤...
and b1≥b2≥b3≥...
But from here on, I don't see why it follows that b_m is an upper bound for the sequence (a_n) for ALL m=1,2,3,... .

Thanks!

 

[iomtt6076]

Thanks, so it just follows from the fact that the least upper bound of (a_n) is less than or equal to any upper bound of (a_n). Got it!:)

One more question...

How can we PROVE that b_m is an upper bound for the sequence (a_n) for ALL m=1,2,3,...?

When I draw a picture, I can see that this is the case, but I am not sure how to write it out and PROVE rigorously.
We know that
a1≤a2≤a3≤...
and b1≥b2≥b3≥...
But from here on, I don't see why it follows that b_m is an upper bound for the sequence (a_n) for ALL m=1,2,3,... .

Thanks!

I think it comes from the hypothesis of nested intervals. If there exists an m such that b_m is not an upper bound, then a_k>b_m for some k. So [a_k,b_k] would be disjoint from [a_m,b_m], which contradicts the assumption of nested intervals. When you have a sequence of nested intervals, no two intervals can be disjoint.

 

[kingwinner]

Makes sense. Thank you! :)

 

[kingwinner]

Another proof:
(a_n) is an increasing sequence bounded above (by b₁), so
lim a_n = sup a_n = a
n->∞

(b_n) is a decreasing sequence bounded below (by a₁), so
lim b_n = inf b_n = b
n->∞

a_n≤b_n for all n
=> a ≤ b

Thus, a_k ≤ a ≤ b ≤ b_k for all k≥1. So the point a belongs to the intersection and the intersection is nonempty.
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Why is the red part true? How can we prove it?

Thanks!

 

[snipez90]

Any of the b_m is an upper bound for the set of all a_n, so the LEAST upper bound a must be less than or equal to b_m for any m. Now you figure out why b must in turn be greater than or equal to a.

 

[kingwinner]

Yes, I already understood that a ≤ b_m for all m. It just follows from the fact that the least upper bound of (a_n) is less than or equal to any upper bound of (a_n). (see post #8)


And now my problem is about the second proof.

How can we prove that "a_n ≤ b_n for all n IMPLIES lim a_n ≤ lim b_n"?

It seems clear in my mind, but how can we prove rigorously using epsilon?

Thanks!

 

[jgens]

If it helps, here's how I might tackle this problem. You already know that $a_n \leq b_m$ for all $m,n \in \mathbb{N}$. From this we can conclude that the sequence $\{a_n\}$ has a least upper bound which we denote $\alpha$. Similarly, the sequence $\{b_n\}$ has a greatest lower bound which we denote $\beta$. Now, from this we have only three possible scenarios:

• $$a_n \leq \alpha < \beta \leq b_n$$

• $$a_n \leq \alpha = \beta \leq b_n$$

• $$a_n \leq \beta < \alpha \leq b_n$$

If either of the first two cases hold, we know that the theorem we're trying to prove is true. If the third case is true however, we know that the intersection of all intervals $[a_n,b_n]$ is necessarily empty. Thus, if you can rule the third case out, then you'll have proved the theorem. So, suppose that the third case is true and let $\alpha - \beta = 2\varepsilon > 0$. Can you derive a contradiction from this?