3.3 * Compact Sets 163
By our hypothesis, n C contains a point of A. But,n C = n {Uc: U EU}= (U Vt.
Thus, (u Vt contains a point of A. But then U U does not contain all of A;
i.e., U is not a cover of A. Contradiction!
Therefore, A is compact. •Applications of compactness will be found at various places later in the
course. See especially Sections 5.3 and 5.4.
EXERCISE SET 3.3l. Prove Theorem 3.3.5.
- Prove Theorem 3.3.6. [Hint: for an unbounded set, try the cover U ={(n-
1, n + 1) : nEN}.] - For each of the sets given in Corollary 3.3.9,
(a) give an open cover that has no finite subcover;
(b) without using open covers, give a simple reason and a theorem showing
that the set is not compact. - Prove that a set of real numbers is bounded iff its closure is compact.
- Prove that the union of finitely many compact sets is compact.
- Prove that the boundary of a bounded set is compact.
- Suppose that {xn} is a convergent sequence; say lim Xn = L. Prove that
n->OO
{xn: n EN} U {L} is compact. - Prove that the intersection of any collection of compact sets is compact.
Is the union of any co llection of compact sets necessarily compact? Justify
your answer. - Prove that every nonempty compact set contains both its supremum and
its infimum. [Thus, it contains both a maximum element and a minimum
element.] - Show that the first part of Cantor's Nested Intervals Theorem (See Al-
ternate 2.5.17) is an easy corollary of Theorem 3.3.17. - If A is a bounded set, we define its diameter to be the real number
d(A) = sup{ Ix -YI : x, y E A}. Prove that if A is compact, then 3
xo, Yo EA 3 d(A) = lxo - Yol·