The Real And Complex Number Systems
The Real And Complex Number Systems
The Real And Complex Number Systems
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exists an n ball Bx, r x such that Bx, r x x a, b . For the point a, it is easy<br />
to know that Ba, r a a, b . That is, in this case, there is only one<br />
accumulation point a of a, b.<br />
So, from 1 and 2, we know that the set of the accumulation points of a, b is a, b.<br />
Since a a, b, we know that a, b cannot contain its all accumulation points. So,<br />
a, b is not closed.<br />
Since an n ball Bb, r is not contained in a, b for any r 0, we know that the point<br />
b is not interior to a, b. So,a, b is not open.<br />
(c) All numbers of the form 1/n, (n 1,2,3,....<br />
Solution: Write the set 1/n : n 1,2,... 1, 1/2, 1/3, . . . , 1/n,... : S.<br />
Obviously, 0 is the only one accumulation point of S. So,S is not closed since S does not<br />
contain the accumulation point 0. Since 1 S, andB1, r is not contained in S for any<br />
r 0, S is not open.<br />
Remark: Every point of 1/n : n 1,2,3,... is isolated.<br />
(d) All rational numbers.<br />
Solutions: Denote all rational numbers by Q. It is trivially seen that the set of<br />
accumulation points is R 1 .<br />
So, Q is not closed. Consider x Q, anyn ball Bx is not contained in Q. Thatis,x<br />
is not an interior point of Q. In fact, every point of Q is not an interior point of Q. So,<br />
Q is not open.<br />
(e) All numbers of the form 2 n 5 m , (m, n 1,2,....<br />
Solution: Write the set<br />
2 n 5 m : m, n 1, 2, . . m m1 1 2 5m , 1 4 5m ,..., 1<br />
2<br />
n 5m ,... : S<br />
1 2 1 5 , 1 2 1 5 ,..., 1 2 2 5 1<br />
m ,... <br />
1 4 1 5 , 1 4 1 5 ,..., 1 2 4 5 1<br />
m ,... <br />
........................................................<br />
<br />
2 1 n 1 5 , 1<br />
2<br />
n 1 5 ,..., 1 2 2<br />
n 5 1<br />
m ... <br />
.....................................................<br />
So,wefindthatS 1 : n 1, 2, . . . 1 : m 1, 2, . . .<br />
2 n 5<br />
0. So,S is not<br />
m<br />
closed since it does not contain 0. Since 1 S, andB 1 , r is not contained in S for any<br />
2<br />
2<br />
r 0, S is not open.<br />
Remark: By (1)-(3), we can regard them as three sequences<br />
1<br />
2 m 5m , 1 5 m m<br />
and 1<br />
m1 4 m1 2<br />
n m 5m , respectively.<br />
m1<br />
<strong>And</strong> it means that for (1), the sequence 5 m m m1<br />
moves 1 . Similarly for others. So, it is<br />
2<br />
easy to see why 1 is an accumulation point of 1 2 2 5m m<br />
m1 . <strong>And</strong> thus get the set of all<br />
accumulation points of 2 n 5 m : m, n 1, 2, . . .<br />
(f) All numbers of the form 1 n 1/m, (m, n 1,2,....<br />
Solution: Write the set of all numbers 1 n 1/m, (m, n 1, 2, . . . as<br />
1 m 1 m m<br />
1 1<br />
m1 m : S.<br />
m1<br />
1<br />
2<br />
3