The Real And Complex Number Systems

Since S R, wehaveS inf S,supS. (Note that we accept that inf S or
sup S .) If S is not an interval, then there exists x inf S,supS such that x S.
(Otherwise, inf S,supS S which implies that S is an interval.) Then we have
1. , x S : A is open in S
2. x, S : B is open in S
3. A B S.
Claim that both A and B are not empty. Asume that A is empty, then every s S, wehave
s x inf S. By the definition of infimum, it is impossible. So, A is not empty. Similarly
for B. Hence, we have proved that S is disconnected, a contradiction. That is, S is an
interval.
Remark: 1. We note that any interval in R is connected. It is immediate from Exercise
4.44. But we give another proof as follows. Suppose there exists an interval S is not
connceted, then there exist two subsets A and B such that
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.
Since A and B , we choose a A and b B, and let a b. Consider
c : supA a, b.
Note that c clA A implies that c B. Hence, we have a c b. In addition,
c B clB, then there exists a B S c; B . Choose
d B S c; c , c S so that
1. c d b and 2. d B.
Then d A. (Otherwise, it contradicts c supA a, b. Note that
d a, b S A B which implies that d A or d B. We reach a contradiction since
d A and d B. Hence, we have proved that any interval in R is connected.
2. Here is an application. Is there a continuous function f : R R such that
fQ Q c ,andfQ c Q
Ans: NO! If such f exists, then both fQ and fQ c are countable. Hence, fR is
countable. In addition, fR is connected. Since fR contains rationals and irrationals, we
know fR is an interval which implies that fR is uncountable, a cotradiction. Hence, such
f does not exist.
4.39 Let X be a connected subset of a metric space S. LetY be a subset of S such that
X Y clX, whereclX is the closure of X. Prove that Y is also connected. In
particular, this shows that clX is connected.
Proof: Given a two valued function f on Y, we know that f is also a two valued
function on X. Hence, f is constant on X, (without loss of generality) say f 0onX.
Consider p Y X, it ,means that p is an accumulation point of X. Then there exists a
dequence x n X such that x n p. Note that fx n 0 for all n. So, we have by
continuity of f on Y,
fp f lim n
x n
lim n
fx n 0.
Hence, we have f is constant 0 on Y. Thatis,Y is conneceted. In particular, clX is
connected.
Remark: Of course, we can use definition of a connected set to show the exercise. But,
it is too tedious to write. However, it is a good practice to use definition to show it. The
reader may give it a try as a challenge.