The Real And Complex Number Systems

(a) intA if A is open or if A is closed in M.
Proof: (1) Suppose that A is open. We prove it by the method of contradiction. Assume
that intA , and thus choose
x intA
intclA clM A
intclA M A
intclA intM A since intS T intS intT.
Since
x intclA Bx, r 1 clA A A
and
x intM A Bx, r 2 M A A c *
we choose r minr 1 , r 2 , then Bx, r A A A c A A c . However,
x A and x A Bx, r A for this r. **
Hence, we get a contradiction since
Bx, r A by (*)
and
Bx, r A by (**).
That is, intA if A is open.
(2) Suppose that A is closed, then we have M A is open. By (1), we have
intM A .
Note that
M A clM A clM M A
clM A clA
A
. Hence, intA if A is closed.
(b) Give an example in which intA M.
Solution: Let M R 1 ,andA Q, then
A clA clM A clQ clQ c R 1 . Hence, we have R 1 intA M.
3.49 If intA intB and if A is closed in M, then intA B .
Proof: Assume that intA B , then choose x intA B, then there exists
Bx, r A B for some r 0. In addition, since intA , we find that Bx, r A.
Hence, Bx, r B A . It implies Bx, r M A . Choose
y Bx, r M A, then we have
y Bx, r By, 1 Bx, r, where 0 1 r
and
y M A By, 2 M A, forsome 2 0.
Choose min 1 , 2 , then we have
By, Bx, r M A
A B A c
B.