The Real And Complex Number Systems

function. Since f0, 1 0, 1 0, 1, we have the domain of f 1 is 0, 1 0, 1 which
is connected. Choose a special point y 0, 1 0, 1 so that f 1 y : x 0, 1.
Consider a continous function g f 1 | 0,10,1y , then
g : 0, 1 0, 1 y 0, x x,1 which is continous. However, it is impossible
since 0, 1 0, 1 y is connected but 0, x x,1 is not connected. So, such f cannot
exist.
Connectedness
4.36 Prove that a metric space S is disconnected if, and only if there is a nonempty
subset A of S, A S, which is both open and closed in S.
Proof: () Suppose that S is disconnected, then there exist two subset A, B in S such
that
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.
Note that since A, B are open in S ,wehaveA S B, B S A are closed in S. So,ifS
is disconnected, then there is a nonempty subset A of S, A S, which is both open and
closed in S.
() Suppose that there is a nonempty subset A of S, A S, which is both open and
closed in S. Then we have S A : B is nonempty and B is open in S. Hence, we have two
sets A, B in S such that
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.
That is, S is disconnected.
4.37 Prove that a metric space S is connected if, and only if the only subsets of S which
are both open and closed in S are empty set and S itself.
Proof: () Suppose that S is connected. If there exists a subset A of S such that
1. A , 2.A is a proper subset of S, 3.A is open and closed in S,
then let B S A, wehave
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.
It is impossible since S is connected. So, this A cannot exist. That is, the only subsets of S
which are both open and closed in S are empty set and S itself.
() Suppose that the only subsets of S which are both open and closed in S are empty
set and S itself. If S is disconnected, then we have two sets A, B in S such that
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.
It contradicts the hypothesis that the only subsets of S which are both open and closed in S
are empty set and S itself.
Hence, we have proved that S is connected if, and only if the only subsets of S which
are both open and closed in S are empty set and S itself.
4.38 Prove that the only connected subsets of R are
(a) the empty set,
(b) sets consisting of a single point, and
(c) intervals (open, closed, half-open, or infinite).
Proof: Let S be a connected subset of R. Denote the symbol #A to be the number of
elements in a set A. We consider three cases as follows. (a) #S 0, (b) #S 1, (c)
#S 1.
For case (a), it means that S , and for case (b), it means that S consists of a single
point. It remains to consider the case (c). Note that since #S 1, we have inf S sup S.