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## Homework Statement

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

(a): If M is any upper bound for A, then: x[itex]\in[/itex](-A), -x[itex]\in[/itex]A, and -x[itex]\leq[/itex]M. Therefore x[itex]\geq[/itex]-M, hence -M is a lower bound for -A. By the Least Upper Bound Property, inf(-A) exists. If inf(-A) exists, then -M[itex]\leq[/itex]inf(-A) for all upper bounds M of the set A. Therefore, -sup(A)[itex]\leq[/itex]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, [itex]\forall[/itex]a[itex]\in[/itex]A and [itex]\forall[/itex]b[itex]\in[/itex]B, there exist a

_{1}, a

_{2}[itex]\in[/itex]A and b

_{1}, b

_{2}[itex]\in[/itex]B such that a

_{1}[itex]\leq[/itex]a[itex]\leq[/itex]a

_{2}and b

_{1}[itex]\leq[/itex]b[itex]\leq[/itex]b

_{2}[itex]\forall[/itex]a[itex]\in[/itex]A and [itex]\forall[/itex]b[itex]\in[/itex]B. Thus (a

_{1}+b

_{1})[itex]\leq[/itex](a+b)[itex]\leq[/itex](a

_{2}+b

_{2}) and (A+B) is also bounded.

Let m=sup(A), n=sup(B). Then for all e>0,

(i)x<m+(e/2) [itex]\forall[/itex]x[itex]\in[/itex]A and y<n+(e/2) [itex]\forall[/itex]y[itex]\in[/itex]B

(ii)x>m-(e/2) for some x[itex]\in[/itex]A and y>n-(e/2) for some y[itex]\in[/itex]B

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

(i) (x+y)<(m+n)+e [itex]\forall[/itex]x[itex]\in[/itex]A, [itex]\forall[/itex]y[itex]\in[/itex]B

(ii) (x+y)>(m+n)-e for some x[itex]\in[/itex]A, for some y[itex]\in[/itex]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.