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**burgow** 1. Let S be a subset of R. Show that S is an inverval if and only if for all a,b in S, we have [a,b] contained in S.

=== I supose the definition of "interval" you have is that an interval is the set of all numbers between two other numbers (extreme points contained or not: open, closed, half open adn etc. intervals)

If so then: If S is an interval the S = {x in IR ; c < s < d } (this is an open interval. The proof is identical if you supose I is closed or whatever).

Now, let a,b in S and assume wrg that a < b ==> c < a < b < d ==> any x s.t. x in [a,b] does ALSO fulfill c < x < d ==> any element in [a,b] is in S.

The other direction: if for any a,b in S then all the elements in [a,b] are in S, then let a = inf S and b = sup S (one or both of these can be -oo or oom we don't care) ==> for any e > 0 there exist elements in S c = a + e, b - e ==> all the elements in [c,d] are in S, and sinc ethis happens for ANY e > 0 we get that S is the interval with extreme points a,b (it can opne, closed of none, according as if we have above supremum or maximum, and infimum or minimum etc.

2. two parts:

prove the following for A,B being bounded sets of the real numbers.

part a: if A+B is defined as the set of all a + b such that a is in A and b is in B

then

LUB(A + B) = LUB(A) + LUB(B)

Supose P = LUB(A), Q = LUN(B) and R = LUB(A+B) ==> for any e > 0 there exist a_e in A and b_e in B s.t. and P - e/2 < a_e <= P, Q - e/2 < b_e <= Q , so P + Q - e = (P - e/2) + (Q - e/2) < a_e + b_e < P + Q (***) , and this is true for ANY e > 0 ==> for ANY e > 0 there exists an element in A + B (namely, a_e + b_e as above) s.t. (***) above is true ==> P + Q = LUB(A + B) by definition.

For the multiplication try to do yourself something symmilar

Tonio

and the second part:

part b: if every element in A,B is positive, and

A*B is defined as the set of all a*b such that a is in A and b is in B

then

LUB(A*B) = LUB(A) * LUB(B)

3. (this one is hard!) (supposed to be done by Pigeon hole)

Let T be a triangle such that T can be placed inside the unit square (sides of length one) with the center of the square not contained in the triangle. Prove that one of the sides of T must have a length of less than 1.

Any ideas? I appreciate the help!