detials

Sol:
Theorem 4.1.20 states that,
If (X, d) is a metric space, then
i. Each connected subset of (X, d) is contained in exactly one component;
ii. Each nonempty connected subset of (X, d) that is both open and closed in (X, d) is a
component of (X, d);
iii. Each component of (X, d) is closed.
We know that,
Union of all components of X forms X itself.
Hence,
* ( ) +
From theorem 4.1.20(i), it can be observed that distinct components are disjoint.
Hence we can say that the connected components of any metric space X forms a partition of X.

Sol:
Given A be a compact subset of the metric space (X, d). Show that for any and is such that
( ) ( )
Let ( ) * ( ) +
For every integer n, there exists and such that ( )



Since A is compact, the sequence * + has a subsequence converging to a point . We claim that
( ) ( )
Suppose ( ) ,
Say ( ) Since a subsequence of * + converges to p, there exists such
that ( )


and ( )



Hence,
( ) ( )



( ) ( )
This is a contradiction to our assumption. Hence ( ) ( )

Suppose ( )
From the above solution, we can say that such that ( ) ( )
Since B is closed and ( ) , it follows that
This means
This contradicts our hypothesis, so ( )

To prove this, it is enough for us to give an example that counters the argument.
Example:
( ) ...