This exercise is meant to illustrate the point made in the o

This exercise is meant to illustrate the point made in the opening paragraph to Section 3.3. Verify that the following three statements are true if every blank is filled in with the word \"finite.\" Which are true if every blank is filled in with the word \"compact.\" Which are true if every blank is filled in with the word \"closed.\" (a) Every set has a maximum. (b) If A and B are, then A + B = {a + b: a A, b B} is also. (c) If {A_n: n N} is a collection of sets with the property that every finite subcollection has a nonempty intersection, then_n = 1^infinity A_n is nonempty as well.

Solution

a)

You have some finite set S and a total order (that is, for any x,ySx, either xy or yx).

Given a non-empty finite set S, we say that m is a maximum of S iff mSmS and for all xS, we have xm.

Induction is straightforward. Suppose S is a non-empty, finite set.

The proposition is that for all sets of size |S|n| that a maximum of S exists.

For n=1, just choose the element of the set.

Suppose the statement is true for n,

let S be a set with |S|=n+1|. Choose xS and form S=S{x}. By assumption, S has a maximum m. If xm, then let m=m, otherwise if mx and mx, let m=x. Then we have mS and xm for all xS, hence m is a maximum of S.

A finite set of real numbers has a maximum and a minimum. Therefore, a finite set is bounded.

Since a finite set has no accumulation points, it follows that a finite set vacuously contains all its accumulation points. Therefore, a finite set is closed.

A closed and bounded set is compact.

b)

Let x A + B. Then x = a + b for some a A and b B.

Let {x_n} be a sequence of numbers belonging to A + B.

Then each x_n = a_n + b_n for some a_n A and b_n B.

The sequence {a_n} has a subsequence {a_nk} that converges to a A since A is compact.

Now the subsequence {b_nk}, being itself a sequence in B, has a converging subsequence {b_nkl} to some b in B, since B is also compact.

This means that the sub(sub)sequence {x_nkl } = {a_nkl} + {b_nkl} converges to a + b A + B.

Hence A + B is compact.