Page 227 - DMTH503_TOPOLOGY

Unit 27: Complete Metric Spaces

27.3 Theorems and Solved Examples Notes

Theorem 2: Let X be a complete metric space and Y be a subspace of X. Show that Y is closed iff it
is complete.

Proof: Let Y be closed.
Let <x > be a Cauchy sequence in Y. This implies that it is a Cauchy sequence in X.
n
Since X is complete, <x > converges to some point xX.
n
Let A be the range of <x >.
n
If A is finite, then x is that term of <x > which is infinitely repeated and therefore xX. If A is
n
infinite, then x, being limit of <x >, is a limit point of its range A. Since AY, so, x is a limit point
n
of Y. But Y is closed, therefore, xY.
This implies that <x > is convergent in Y. Hence Y is complete.
n
Conversely, let Y be complete.

Here we are to prove that Y is closed.
Let x be a limit point of Y.

æ 1 ö
Then, for each positive integer n,an open sphere S x, n ø ÷ containing at least one point x of Y,
ç
n
è
other than x.
Let> 0 be given.
1 1
a positive integer n such that < . We have <  for all nn .
o o
n n
o
æ 1 ö
Since x  S x, ÷ ,
ç
n è n ø
1
d(x , x) < .
n
n
Therefore d(x , x) <  n  n .
n o
This implies that <x > converges to x in X. Therefore <x > is a Cauchy sequence in X, So it is a
n n
Cauchy sequence in Y.
But Y is complete.

Therefore <x > is convergent in Y.
n
This implies that x  Y, because limit of convergent sequence is unique. Hence, Y is closed.
Theorem 3: Cantor’s Intersection Theorem.

Let X be a complete metric space. Let {F } be a decreasing sequence of non-empty closed subsets
n
¥
of X such that d(F ) ® 0 as n ® ¥. Then F contains exactly one point.
n n
=
n 1
¥
Proof: Let F = F .
n
=
n 1
For nN, let x  F , we prove that <x > is a Cauchy sequence.
n n n
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