The Real And Complex Number Systems

Hence from (1)-(4), we know that isametricforS 1 S 2 .
For other metrics, we define
1 x, y : d 1 x 1 , y 1 d 2 x 2 , y 2 for , 0.
d
2 x, y : d 1 x 1 , y 1 2 x 2 , y 2
1 d 2 x 2 , y 2
and so on. (The proof is similar with us by above exercises.)
Compact subsets of a metric space
3.38 Assume S T M. Then S is compact in M, d if, and only if, S is compact in
the metric subspace T, d.
Proof: Suppose that S is compact in M, d. LetF O : O is open in T be an
open covering of S. SinceO is open in T, there exists the corresponding G which is open
in M such that G T O . It is clear that G forms an open covering of S. Sothereis
a finite subcovering G 1 ,...,G n of S since S is compact in M, d. Thatis,S kn k1 G k .
It implies that
S T S
T kn k1 G k .
kn k1 T G k
kn k1 O k F.
So, we find a fnite subcovering O 1 ,...,O n of S. Thatis,S is compact in T, d.
Suppose that S is compact in T, d. LetG G : G is open in M be an open
covering of S. SinceG T : O is open in T, the collection O forms an open
covering of S. So, there is a finite subcovering O 1 ,...,O n of S since S is compact in
T, d. Thatis,S kn k1 O k . It implis that
S kn k1 O k kn k1 G k .
So, we find a finite subcovering G 1 ,...,G n of S. Thatis,S is compact in M, d.
Remark: The exercise tells us one thing that the property of compact is not changed,
but we should note the property of being open may be changed. For example, in the
2 dimensional Euclidean space, an open interval a, b is not open since a, b cannot
contain any 2 ball.
3.39 If S is a closed and T is compact, then S T is compact.
Proof: Since T is compact, T is closed. We have S T is closed. Since S T T, by
Theorem 3.39, we know that S T is compact.
3.40 The intersection of an arbitrary collection of compact subsets of M is compact.
Proof: Let F T : T is compacet in M , and thus consider TF
T, whereF F.
We have TF
T is closed. Choose S F . then we have TF
T S. Hence, by
Theorem 3.39 TF
T is compact.
3.41 The union of a finite number of compact subsets of M is cmpact.
Proof: Denote T k is a compact subset of M : k 1, 2, . . n by S. LetF be an open
covering of kn k1 T k . If there does NOT exist a finite subcovering of kn k1 T k , then there
does not exist a finite subcovering of T m for some T m S. SinceF is also an open
covering of T m , it leads us to get T m is not compact which is absurb. Hence, if F is an open
covering of kn k1 T k , then there exists a finite subcovering of kn k1 T k .So,kn k1 T k is