Proving the Greatest Lower Bound Property with

[major_maths]

Homework Statement

Use part (a) to prove the Greatest Lower Bound Property.

(a): If M is any upper bound for A, then: x$\in$(-A), -x$\in$A, and -x$\leq$M. Therefore x$\geq$-M, hence -M is a lower bound for -A. By the Least Upper Bound Property, inf(-A) exists. If inf(-A) exists, then -M$\leq$inf(-A) for all upper bounds M of the set A. Therefore, -sup(A)$\leq$inf(-A).

If -sup(A)<inf(-A), then it follows that there exists some distance between -sup(A) and inf(-A). Assuming set A is reflected correctly when transformed into set -A, it would be necessary for there to be a difference in the amount of elements in sets A and -A to create this distance. Set A must contain the same amount of elements as set -A by definition. Therefore, -sup(A) = inf(-A).

Since A and B are bounded, $\forall$a$\in$A and $\forall$b$\in$B, there exist a₁, a₂$\in$A and b₁, b₂$\in$B such that a₁$\leq$a$\leq$a₂ and b₁$\leq$b$\leq$b₂ $\forall$a$\in$A and $\forall$b$\in$B. Thus (a₁+b₁)$\leq$(a+b)$\leq$(a₂+b₂) and (A+B) is also bounded.

Let m=sup(A), n=sup(B). Then for all e>0,
(i)x<m+(e/2) $\forall$x$\in$A and y<n+(e/2) $\forall$y$\in$B
(ii)x>m-(e/2) for some x$\in$A and y>n-(e/2) for some y$\in$B

[Combining the inequalities in each (i) and (ii)]

(i) (x+y)<(m+n)+e $\forall$x$\in$A, $\forall$y$\in$B
(ii) (x+y)>(m+n)-e for some x$\in$A, for some y$\in$B

Hence, sup(A+B) = (m+n) = sup(A) + sup(B).

Homework Equations

See above.

The Attempt at a Solution

I haven't the slightest clue how to start this proof, so if you could give me that, I should be able to work out the rest of it.

 

[micromass]

You need to prove that each bounded-below set A has an infimum. Can you prove that -A has a supremum, first?