The Real And Complex Number Systems

Bx, r x S Bx, r x T .
So, at least one of Bx, r x S and Bx, r x T is not empty. If
Bx, r x S , then x S .AndifBx, r x T , then x T .So,
S T S T .
From 1 and 2, we have S T S T .
Remark: Note that since S T S T ,wehaveclS T clS clT,
where clS is the closure of S.
(d) S S .
Proof: Since S S S , then S S S S S S since S S by
(a).
(e) S is closed in R n .
Proof: Since S S S by (d), then S cantains all its accumulation points. Hence, S
is closed.
Remark: There is another proof which is like (a). But it is too tedious to write.
(f) S is the intersection of all closed subsets of R n containing S. That is, S is the
smallest closed set containing S.
Proof: It suffices to show that S AS A,whereA is closed. To show the statement,
we consider two steps as follows.
1. Since S is closed and S S, then AS A S.
2. Let x S, then Bx, r S for any r 0. So, if A S, then
Bx, r A for any r 0. It implies that x is an adherent point of A. Hence if A S,
and A is closed, we have x A. Thatis,x AS A.So,S AS A.
From 1 and 2, we have S AS A.
Let S T S, whereT is closed. Then S AS A T. It leads us to get T S.
That is, S is the smallest closed set containing S.
Remark: In the exercise, there has something to remeber. We list them below.
Remark 1. If S T, then S T .
2. If S T, then S T.
3. S S S .
4. S is closed if, and only if S S.
5. S is closed.
6. S is the smallest closed set containing S.
3.13 Let S and T be subsets of R n . Prove that clS T clS clT and that
S clT clS T if S is open, where clS is the closure of S.
Proof: Since S T S and S T T, then clS T clS and,
clS T clT. So,clS T clS clT.
Given an open set S , and let x S clT, then we have