Page 238 - DMTH503_TOPOLOGY

Topology

Notes Theorem 5: Every compact subset of a metric space is closed and bounded.
Proof: Let Y be a compact subset of a metric space (X, d). If f is finite, then it is certainly bounded
and closed.

Consider the case in which Y is not finite.
Y is compact  Y is sequentially compact.
To prove that Y is bounded. Suppose not. Then Y is not bounded. Then it is possible to find a pair
of points of Y at large distance apart. Let y  Y be arbitrary.
1
Then we take y  Y
2
s.t. d(y , y ) > 1
1 2
Now we can select a point y s.t.
3
d(y , y ) > l + d (y , y )
1 3 1 2
Continuing this process, we get a sequence
y   Y
n
with the property that d(y , y ) > 1 + d(y , y m–1  n  N
)
1 m 1
 d(y , y ) > 1 + d(y , y ) for m > n (1)
1 m 1 n
This d(y , y )  d(y , y ) – d(y , y ) > 1
n
1
m
1
m n
Above relation shows that y  has no convergent subsequence contrary to the fact that Y is
n
sequentially compact. Hence Y is bounded.
Aim: Y is closed.
Let y be a limit point of Y,  sequence
y   Y s.t. lim y = y
n n
Every sequence of y  converges to y. For Y is sequentially compact and so every sequence in Y
n
must converge in Y.
Hence y  Y
Thus y  D(Y)  y  Y
or D(Y)  Y or Y is closed.
Theorem 6: Every sequentially compact metric space is compact.
Proof: Let (X, d) be a sequentially compact metric space. To prove that X is compact.

Since X is sequentially compact metric space.
X is totally bounded. Let  > 0 be an arbitrary real number fixed.
X is totally bounded  X has  - net.

Let us denote the set  net by A.
Then A is finite subset of X with the property
X = {S : a  A} ...(1)
(a)
Since A is finite and hence we can write
A = {x , x , x ,....., x }
1 2 3 n

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