# LUB and Nested Interval Equivalancy

1. Dec 8, 2004

### dogma

Hello there.

I understand (to the best of my ability) the Least Upper Bound property and the Nested Interval property, but I don't see how the two are equivalent properties.

LUB:

If $$S \subset \mathbb{R}$$ has an upper bound, then $$S$$ has a LUB

Nested Intervals:

If $$I_1 \supset I_2 \supset \cdots \supset I_n \supset \cdots$$ is a sequence of nested, closed, bounded, non-empty intervals, then

$$\bigcap_{n=1}^{\infty}I_n \neq \emptyset$$

and length($$I_n \rightarrow 0$$, then $$\exists$$

$$x_o \in \bigcap_{n=1}^{\infty}I_n$$

(ie: a unique point exists in all the intervals)

Thanks in advance for clarifying this for me.

dogma

Last edited: Dec 8, 2004
2. Dec 8, 2004

### HallsofIvy

Staff Emeritus
First, the "least upper bound property" is equivalent to the "greatest lower bound property" (If a set of real numbers has a lower bound, then it has a greatest lower bound). If X is a set having a lower bound b, -X (multiply each member of X by -1) has upper bound -b and so has a least upper bound, x. -x is then the greatest lower bound of X.

The left endpoints of the intervals form a set having all the right endpoints as upper bounds therefore, has a least upper bound, a. The set of right endoints has a as lower bound therefore has a greatest lower bound, b. Thus: a<= b so the interval [a,b] is in every interval. If length In goes to 0, a must equal b since otherwise there would be a In with length smaller than b-a.

I'm going to have to think about proving the other way!

3. Dec 8, 2004

### dogma

thanks...that helps clear up things alot.

I'm still having a hard time seeing the other way...I'm sure it's not as bad as I think.

dogma

4. Dec 8, 2004

### Hurkyl

Staff Emeritus
My book diagrams a circuitous chart of implications that prove the equivalency of various properties. The chain it depicts for this is:

Nested Interval Property --> Bolzano-Wierstrass theorem --> Monotonic sequence property --> LUB property (--> Heine-Borel theorem --> Nested Interval Property)

Dedikend completness, connectedness of the line, Archimedian property, and convergence of Cauchy sequences are also on the chart. (The last two alone aren't enough to prove the others, but together they are)

I'll just give hints on the implications, I don't want to spoil it. In fact, you might not want to read ahead!

You prove the Bolzano-Wierstrass theorem by starting with a closed interval containing your set. You then split the interval in half...

I don't think you need a hint to prove the monotonic sequence property.

To prove the LUB property, start with two numbers, one an upper bound, and one not an upper bound. Now, consider their midpoint...

Last edited: Dec 8, 2004
5. Dec 9, 2004

### HallsofIvy

Staff Emeritus
I can prove that the "nested interval" property implies the "compactness property" (every closed and bounded set is compact) which then implies LUB.

Lemma 1: If A is compact and B is a closed subset of A then B is compact.
Proof: If Un is a an open cover for B then adding the complement of B (which is open since B is closed) gives a open cover for A. Since A is compact, there is a finite sub-cover of A. Dropping the complement of B from that gives an open sub-cover of B.

Lemma 2: The closed bounded interval [a, b] is compact.
Suppose it is not. Then there exist an open cover which has no finite sub-cover. Take c to be the mid-point of [a,b]. Then either [a, c] or [c, b] (or both) cannot be covered by a finite subset of the open cover. Divide that interval in half- again, at least one of the "halves cannot be covered by a finite subset of the open cover. Continuing in this way, we get a nested sequence of intervals, of length (b-a)/22, none of which can be covered by a finite subcover. BY THE NESTED INTERVAL PROPERTY, there is a unique point, p, in all the intervals. At least one of the sets, call it U, in the original open cover contains p. Since that set is open, there is some interval, [p-delta, p+delta] in U. But since the length of the intervals goes to 0, some interval has length less than delta and so is completely contained in U which contradicts the fact that no finite subcover would cover any of the intervals.

Theorem: Every closed and bounded set is compact.
Suppose A is closed and bounded. Because A is bounded, it is contained in some interval [a, b] which, by lemma 2 is compact. Because A is closed, it is compact by lemma 2.

"Compactness property" implies LUB
We will prove the contrapositive: if LUB is not true then the compactness property is not true.

Suppose there exist a non-empty set, A, having upper bound b, but no least upper bound. Since A is non-empty, there exist x0 in A. Since A has no least upper bound, it does not have a largest member and so there also exist x1in A with x0< x1. Let X= [x1,b]U closure(A). X is both closed (because it is the union of two open sets) and bounded (because it is contained in [x1,b]). We will show that X is not compact by exhibiting an open cover that has no finite sub-cover.

For any x contained in A, x> 1, let Ux be the open interval (x0, x). If p is in X, p> x1> x0. To show that p is in Ux for some x we need only show that there exist x in A such that p< x. p is either in A or is a limit point of A.
If p is in A then, as before, since A has no largest member then there exist x in A such that p< x.
If p is not in A then it is a limit point of A: for any delta, there exist x in A such that x is in (p- delta, p+ delta). If there were no x in A larger than p (so that p is an upper bound on A), then we have x in (p- delta,delta) which say that p is the least upper bound of A- which contradicts the hypothesis that A has no LUB.
That is, the collection {Ux} is an open cover for A.

But there clearly is no finite sub-cover: since all sets are of the form (x0, x), any finite number of them has a largest such x. That x itself is in X yet is not in any of the intervals in the finite collection. QED.

6. Dec 11, 2004

### dogma

A big, belated thanks to all. I apologize for not thanking you all earlier...got caught up with work stuff.

Thanks again for the help.

dogma