Page 228 - DMTH503_TOPOLOGY

Topology

Notes Let> 0 be given.
d(F )®0, therefore there exists a positive integer n of such that d(F ) < .
n o n o
Since <F > is a decreasing sequence,
n
 m, nn  F , F F
o m n n o
x , x F
m n n o
d(x , x ) < d(F )
m n n o
d(x , x ) <  [ d(F ) < ]
m n n o
 <x > is a Cauchy sequence.
n
Since the space X is complete, <x > must converge to some point, say x in X i.e. x ®xX.
n n
We shall prove that
¥
x  F .
n
n 1
=
¥
If possible, let x F .
n
=
n 1
 xF for some kN.
k
Since each F is a closed set, F is also a closed set, therefore x cannot be a cluster point of F , and
n k k
so d(x, F )0.
k
Let d(x, F ) = r > o so that
k
d(x, y)  r  y  F .
k
æ 1 ö
This shows that F  S x, r ÷ = .
ç
k è 2 ø
Now, n > k  F F
n k
 x F ( x F F )
n k n n k
æ 1 ö æ 1 ö
 x  S x, r ÷ [ F  S x, r ÷ = ]
ç
ç
n è 2 ø k è 2 ø
This contradicts the fact that x ®x.
n
¥ ¥
Therefore x F and hence F .
n
n
n 1 n 1
=
=
Example 3: Show that every compact metric is complete.
Solution: Let (X, d) be a compact metric space.
To prove : X is complete.
Let <a > be an arbitrary Cauchy sequence in X. If we show that <a > converges to a point in X, the
n n
result will follow.
X is compact  X is sequentially compact.
 Every sequence in X has a convergent subsequence.
 In particular, every Cauchy sequence in X has a convergent subsequence.
 <a > has a subsequence <a : n  N> which converges to a point a  X
n in io
 <a > also converges to the point a  X.
n i o

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