The Real And Complex Number Systems

we have intS T AS A which implies intS T by intS AS A. Hence, intS is the
largest open subset of S.
3.11 If S and T are subsets of R n , prove that
intS intT intS T and intS intT intS T.
Proof: For the part intS intT intS T, we consider two steps as follows.
1. Since intS S and intT T, wehaveintS intT S T which implies
that Note that intS intT is open.
intS intT intintS intT intS T.
2. Since S T S and S T T, wehaveintS T intS and
intS T intT. So,
intS T intS intT.
From 1 and 2, we know that intS intT intS T.
For the part intS intT intS T, we consider intS S and intT T. So,
intS intT S T
which implies that Note that intS intT is open.
intintS intT intS intT intS T.
Remark: It is not necessary that intS intT intS T. For example, let S Q,
and T Q c , then intS , andintT . However, intS T intR 1 R.
3.12 Let S denote the derived set and S theclosureofasetSinR n . Prove that
(a) S is closed in R n ; that is S S .
proof: Let x be an adherent point of S . In order to show S is closed, it suffices to
show that x is an accumulation point of S. Assume x is not an accumulation point of S, i.e.,
there exists d 0 such that
Bx, d x S . *
Since x adheres to S , then Bx, d S . So, there exists y Bx, d such that y is an
accumulation point of S. Note that x y, by assumption. Choose a smaller radius d so that
By, d Bx, d x and By, d S .
It implies
By, d S Bx, d x S by (*)
which is absurb. So, x is an accumulation point of S. Thatis,S contains all its adherent
points. Hence S is closed.
(b) If S T, then S T .
Proof: Let x S , then Bx, r x S for any r 0. It implies that
Bx, r x T for any r 0sinceS T. Hence, x is an accumulation point of T.
That is, x T .So,S T .
(c) S T S T
Proof: For the part S T S T , we show it by two steps.
1. Since S S T and T S T, wehaveS S T and T S T by (b).
So,
S T S T
2. Let x S T , then Bx, r x S T . Thatis,