> 0, there
is a natural number N such that d(xn , xm) <
if N 
n,m.
Defn
Suppose (X,d)
is a metric space. A sequence {xn} is said to a be Cauchy
sequence in (X,d) if, for
each > 0, there
is a natural number N such that d(xn , xm) <
if N n,m.
Proposition Convergent sequences are Cauchy.
Lemma Cauchy sequences are bounded, but not necessarily convergent.
Pf: Consider X as the interval (0,1] with
the absolute value as norm. The sequence {xn} with xn
= 1/n is Cauchy but not convergent in X. To show that each Cauchy sequence
is bounded, apply the definition with :=
1 to obtain an N such that d(xn , xm)
< 1 if N n,m.
Let R := max {1, d(xN,
x1), d(xN, x2), ...,d(xN, xN-1)},
then {xn} is contained in B2R(xN).
Defn If (X,d) is a metric space for which each Cauchy sequence converges, then (X,d) is said to be a complete metric space.
Lemma If a subsequence of a Cauchy sequence converges, then the sequence is itself convergent to the same limit.
Proposition If C is a closed subset of a complete metric space (X,d), then C is a complete metric space with the restricted metric.
Examples R,
C, Rk, Ck are all complete metric spaces.
Pf: Use the fact that convergence of a sequence in each of the spaces C,
Rk, Ck is equivalent to convergence in each coordinate.of
the sequence. This follows since the sup norm is equivalent to the Euclidean
norm.
Theorem Let C[a,b] denote the normed linear space of continuous functions on the interval [a,b] equipped (as before) with the sup-norm, then C[a,b] is complete.
Pf: Let {fn} be a Cauchy sequence in C[a,b].
Step 1 Establish a pointwise limit for {fn}and call this
function f.
Step 2: Prove that || fn - f ||
0.
Step 3: Next show that the function f is continuous on
[a,b]. (Hint: Use an /3 argument.)