Page 226 - DMTH503_TOPOLOGY

Topology

Notes Choose K = max. (n , m )
o o o
(b , p)  (b , b ) + (b , a ) + (a , p)
n n m m m m
< K + K + K = 3 K m, n  K .
o
1
Choosing initially K = , we get
3
(b , p) <  n  K
n o
This b  p.
n
Conversely if b  p, then by making parallel arguments, we can show that a  p. Hence the
n n
result.
Self Assessment

1. In any metric space, prove that every Cauchy sequence is totally bounded.
2. Let a subsequence of a sequence <a > converge to a point p. Prove that <a > also converges
n n
to p.
27.2 Complete Metric Space

A metric space X is said to be complete if every Cauchy sequence of points in X converges to a
point in X.

Example 2: The complex plane C is complete.
Solution: Let <z > be a Cauchy sequence of complex numbers, where Z = x + i y .
n n n n
Here <x > and <y > are themselves Cauchy sequences of real numbers,
n n
x - x  z - z n
n
m
m
and y - y  z - z
m n m n
But the real line being a complete metric space, there exists real numbers x and y such that x  x
n
and y y.
n
Thus, taking z = x + iy, we find z z as
n
-
+
z - z = (x + i y ) (x i y)
n n n
+
= (x - x) i(y - y)
n
n
 x - x + y - y
n n
 0 as n
 z - z = 0z z.
n n
Hence if the real line is a complete metric space, then the complex plane is also a complete metric
space.

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