This exercise uses the notation from Exercise 12. (a) Let S be a set of numbers

Question

This exercise uses the notation from Exercise 12.

(a) Let S be a set of numbers that is bounded from below by a number B. Show that ?S is bounded from above by ?B.

(b) Let S be as in part (a). It follows from (a) that ?S is bounded from above. The Least Upper Bound Theorem implies that ?S has a least upper bound L. Prove that ?L is the GLB for S. For this you must explain why (i) ?L is a lower bound for S and (ii) why there is no greater lower bound.

Remark It follows from (b) that S has a GLB. Hence the above sequence of arguments proves the GLB property.

(12) If S is any set of numbers, we define ?S = {?x|x ? S}

Compute the sup and inf of ?S for each set S in Exercise 9.

HowdothesupandinfofS relatetothesupandinfof?S?

Explanation / Answer

a)

Suppose ?K?R:?x?S:|x|?K.

Then by Negative of Absolute Value, ?x?S:?K?x?K.

Thus by definition, S is bounded both above and below, and is therefore bounded.

Now suppose S is bounded.

Then by definition S is bounded both above and below.

As S is bounded below, ?L?R:?x?S:L?x.

As S is bounded above, ?H?R:?x?S:H?x.

Now let K=max{|L|,|H|}.

Then K?|L| and K?|H|.

It follows from Negative of Absolute Value that ?K?L?K and ?K?H?K.

In particular:

Thus ?x?S:?K?x?K and, by Negative of Absolute Value, ?x?S:|x|?K.

S is bounded below, so ?m?Z:?s?S:m?s.

Hence ?s?S:0?s?m.

Thus the set T={s?m:s?S}?N.

The Well-Ordering Principle gives that T has a smallest element, which we can call bT?T.

Hence (?s?S:bT?s?m)?(?bS?S:bT=bS?m).

So:

So bS is the smallest element of S.

s?S:bS?m?s?m
? ?s?S:bS?s Cancellability of elements of Z
? bS?S?(?s?S:bS?s) ... which is how smallest element is defined.

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