Defn. Suppose that K Í IR.   A collection G of open subsets such that

Defn. K is called compact, if each open cover  G  of K has a finite subcover.

Theorem. The continuous image of a compact set is compact.
 

Pf.  Suppose  f: K ® IR is continuous and K is compact.  Each open cover  C of f[K] can be drawn back to an open cover  C^~ of K, by considering the sets 
       O^~; = f ^-1 ( O),   OΠ C
K compact implies that we may draw a finite subcover from { C^~}. Each of these members is the inverse image (under f) from a member of  C. These form the desired subcover of f[K].  ^[¯]

Theorem. (Heine-Borel) Suppose that a £ b, then the interval [a,b] is compact.
 

Pf. Let  C be an open cover for [a,b] and consider the set

Note that A ¹Æ since a Î A. Let g: = lub (A). It is enough to show that g > b, since if x₁ Î A and a £ x £ x₁, then x Î A. Suppose instead that g £ b, then there must be some  O₀ Π C such that g Π O₀. But O₀ is open, so there exists d > 0 so that N_d(g) Í  O₀. Since g is the least upper bound for A, then there is an x Î A such that g-d < x £ g. But x Î A implies there are members O₁, ..., O_n of  C whose union covers [a,x]. The collection O₀, O₁, ..., O_n covers [a,g+d/2]. Contradiction, since g is the least upper bound for the set A.   ^[¯]

Theorem. Each closed subset C of a compact set K is compact.
 

Pf. Let G^~ be an open cover for C. Let  O₀ be the complement of C, then  O₀ is open and  G: = G^~ È{ O₀} is an open cover for K.  There is a finite subcover of G which covers K and hence C. This subcover (dropping  O₀ if it appears) is the desired finite subcover for C.   ^[¯]

Defn. Suppose {a_n } is a sequence. A sequence {b_k } is called a subsequence of {a_n } if there exists a strictly increasing sequence of natural numbers

Theorem. Suppose that K Í IR, then TFAE:

a.) K is compact,

b.) K is closed and bounded,

c.) each sequence in K has a subsequence which converges to a member of K,

d.) (Bolzanno-Weierstrass) each infinite subset of K has a limit point in K.

Pf. (a)Þ (b): To show that K is bounded, consider the open cover of K consisting of the collection of nested open intervals  O_n : = (-n,n),  n Î IN. To show that K is closed, let x₀ be a limit point of K. Assume to the contrary that x₀\not Î K. Consider the open cover of K consisting of the collection of nested open sets O_n : = {x Î IR||x-x₀| > 1/n },  n Î IN. Any finite subcollection which would cover K would have union whose complement would be a neighborhood of x₀ not intersecting K. This shows that x₀ could not be a limit point of K.

(b)Þ (d): We use the `divide and conquer' method, better known as the `bisection' method. Let A be an infinite subset of K. Since K is bounded, there is an interval [a,b] such that K Í [a,b]. Inductively define the closed subintervals as follows. Let [a₀,b₀]: = [a,b]. Either the left or right half of [a₀,b₀] contains an infinite number of members of K. In the case that it is the right half, set [a₁,b₁]: = [(b₀+a₀)/2,b₀]. Set [a₁,b₁] equal to the left half of [a₀,b₀] otherwise. Inductively, let [a_n+1,b_n+1] be the half of [a_n,b_n] which contains an infinite number of members of A. Notice that the length of this interval is (b-a)/2ⁿ⁺¹, that the a_n's satisfy a_n £ a_n+1 £ ... < b and so must converge to some real number a £ x₀ £ b. Each neigborhood of x₀ will contain one of the intervals [a_n,b_n] and hence will contain an infinite number of members of A, i.e. x₀ is a limit point of A. This also shows that x₀ is a limit point of the closed set K and must therefore belong to K.

(d)Þ (c): Let { x_n}_{n = 1}^¥ be a sequence in K. If the sequence's image is finite, then we may construct a constant subsequence which has the value which we may choose as any of the values of { x_n}_{n = 1}^¥ which is repeated infinitely often. Otherwise, let A be the range of the sequence. Then A is an infinite subset of K. By the Bolzanno-Weierstrass property, A must have a limit point (x₀ say) which belongs to K. For each k Î IN, we may find an integer n_k larger than those previously picked (i.e., n₁, ...,n_k-1), so that |x_nk-x₀| < 1/k. This is the desired subsequence.

(c)Þ (b): If K were not bounded, then there would exist a sequence x_nÎ K such that |x_n| > n. If this sequence had a subsequence which converged, then it would have to be bounded. But each subsequence of { x_n} is clearly unbounded. To show that K is closed, we let x₀ be a limit point of K which is not in K. We can then find a sequence { x_n} from K which converges to x₀. By condition (c), this has to have a subsequence which converges to a member of K. Contradiction. Each subsequence of a convergent sequence converges to the same limit, in this case x₀, which does not belong to K.   ^[¯]

Corollary. Each continuous function f on a compact set K is bounded.
 

Pf. The set f(K) is compact and is therefore bounded.   ^[¯]

Corollary. (Extreme Value Theorem) Each continuous function on a compact set attains its maximum (resp. minimum).
 

Pf. The set f(K) is compact and is therefore bounded and closed. Hence the least upper bound g for f(K) must belong to f(K). Therefore, there is an x₀ Î K such that g = f(x₀) and so
        |f(x) £ f(x₀),  for all  x Î K.
Similary, the greatest lower bound of f(K) is attained by some member of K.   ^[¯]

Defn. A function f is called uniformly continuous if for each e > 0, $d > 0 such that whenever x₁, x₂Î dom(f) and |x₁ - x₂| < d, then |f(x₁)-f(x₂)|< e.

Corollary. Each continuous function on [a,b] is uniformly continuous.
 

Pf. Suppose not, then negating the definition implies that there exist an e₀> 0 such that for each n Î IN we can find x_n, y_n Î K with |x_n-y_n| < 1/n but |f(x_n) -f(y_n)|³e₀. K is compact so we can find a subsequence { x_nk }_{k = 1}^¥ of { x_n }_{n = 1}^¥ which converges to some x₀ belonging to K. Notice that { y_nk}_{k = 1}^¥ also converges to x₀ (use an e/2 proof). But f is continuous at x₀, so

which is a contradiction.  ^[¯]