Suppose E notequalto 0 is sequentially compact Prove that su

Suppose E notequalto 0 is sequentially compact. Prove that sup(E) is in E. (Similarly, and you need not write down the details, inf (E) is in E.)

Solution:

We will show that sup E is contained in E.

We know that an unbounded set cannot be compact (the cover Gn = (n, n) suffices to show this).

Thus E has an upper bound, and hence has a finite supremum, which we shall call s = sup E.

Suppose for contradiction that s E, and consider the cover Gn = (-,s-1/n).

The union of all these sets covers (, s), which, since s is an upper bound for E, covers E.

But no finite subcover suffices to cover E. This is because any finite subcover has a maximum index N , and so only covers (-,s-1/N).

If this sufficed to cover E, then s-1/N would be an upper bound for E, contradicting that s is the least upper bound. Thus s E implies E not compact.

Thus the result follows.