The Real And Complex Number Systems

That is, intB which is absurb. Hence, we have intA B .
3.50 Give an example in which intA intB but intA B M.
Solution: Consider the Euclidean sapce R 1 , |. |. LetA Q, andB Q c , then
intA intB but intA B R 1 .
3.51 A clA clM A and A M A.
Proof: By the definition of the boundary of a set, it is clear that
A clA clM A. In addition, A clA clM A, and
M A clM A clM M A clM A clA. Hence, we have
A M A.
Remark: It had better regard the exercise as a formula.
3.52 If clA clB , then A B A B.
Proof: We prove it by two steps.
(1) Let x A B, then for all r 0,
Bx, r A B Bx, r A Bx, r B
and
Bx, r A B c Bx, r A c B c *
Note that at least one of Bx, r A and Bx, r B is not empty. Without loss of
generality, we say Bx, r A . Then by (*), we have for all r 0,
Bx, r A , andBx, r A c .
That is, x A. Hence, we have proved A B A B.
(2) Let x A B. Without loss of generality, we let x A. Then
Bx, r A , andBx, r A c .
Since Bx, r A , wehave
Bx, r A B Bx, r A Bx, r B . **
Claim that Bx, r A B c Bx, r A c B c . Suppsoe NOT, it means that
Bx, r A c B c . Then we have
Bx, r A Bx, r clA
and
Bx, r B Bx, r clB.
It implies that by hypothesis, Bx, r clA clB which is absurb. Hence, we have
proved the claim. We have proved that
Bx, r A B by(**).
and
Bx, r A B c .
That is, x A B. Hence, we have proved A B A B.
From (1) and (2), we have proved that A B A B.
Supplement on a separable metric space
Definition (Base) A collection V of open subsets of X is said to be a base for X if the
following is true: For every x X and every open set G X such that x G, we
have
x V G for some .