The Real And Complex Number Systems

compact.
3.42 Consider the metric space Q of rational numbers with the Euclidean metric of
R 1 . Let S consists of all rational numbers in the open interval a, b, whereaandbare
irrational. Then S is a closed and bounded subset of Q which is not compact.
Proof: Obviously, S is bounded. Let x Q S, then x a, orx b. Ifx a, then
B Q x, d x d, x d Q Q S, whered a x. Similarly, x b. Hence, x is an
interior point of Q S. Thatis,Q S is open, or equivalently, S is closed.
Remark: 1. The exercise tells us an counterexample about that in a metric space, a
closed and bounded subset is not necessary to be compact.
2. Here is another counterexample. Let M be an infinite set, and thus consider the
metric space M, d with discrete metric d. ThenbythefactBx,1/2 x for any x M,
we know that F Bx,1/2 : x M forms an open covering of M. It is clear that there
does not exist a finite subcovering of M. Hence, M is not compact.
3.In any metric space M, d, we have three equivalent conditions on compact which
list them below. Let S M.
(a) Given any open covering of S, there exists a finite subcovering of S.
(b) Every infinite subset of S has an accumulation point in S.
(c) S is totally bounded and complete.
4. It should be note that if we consider the Euclidean spaceR n , d, wehavefour
equivalent conditions on compact which list them below. Let S R n .
Remark (a) Given any open covering of S, there exists a finite subcovering of S.
(b) Every infinite subset of S has an accumulation point in S.
(c) S is totally bounded and complete.
(d) S is bounded and closed.
5. The concept of compact is familar with us since it can be regarded as a extension of
Bolzano Weierstrass Theorem.
Miscellaneous properties of the interior and the boundary
If A and B denote arbitrary subsets of a metric space M, prove that:
3.43 intA M clM A.
Proof: In order to show the statement, it suffices to show that M intA clM A.
1. Let x M intA, we want to show that x clM A, i.e.,
Bx, r M A for all r 0. Suppose Bx, d M A for some d 0. Then
Bx, d A which implies that x intA. It leads us to get a conradiction since
x M intA. Hence, if x M intA, then x clM A. Thatis,
M intA clM A.
2. Let x clM A, we want to show that x M intA, i.e., x is not an interior
point of A. Suppose x is an interior point of A, then Bx, d A for some d 0. However,
since x clM A, then Bx, d M A . It leads us to get a conradiction since
Bx, d A. Hence, if x clM A, then x M intA. Thatis,clM A M intA.
From 1 and 2, we know that M intA clM A, or equvilantly,
intA M clM A.