The Real And Complex Number Systems

for every real number y, either there is no x in 0, 1 for which fx y or
there are exactly two values of x in 0, 1 for which fx y
Assume that there exist a finite many numbers of discontinuities of f, say these points
x 1 ,...,x n . By property A, there exists a unique y i such that fx i fy i for 1 i n.
Note that the number of the set
S : x 1 ,...,x n y 1 ,...,y n x : fx f0, andfx f1 is even, say 2m
We remove these points from S, and thus we have 2m 1 subintervals, say I j ,
1 j 2m 1. Consider the local extremum in every I j ,1 j 2m 1 and note that
every subinterval I j ,1 j 2m 1, has at most finite many numbers of local extremum,
say # t I j : fx is the local extremum t j j
1 ,..,t pj p j . And by property A,
there exists a unique s j j
j
k such that f t k f s k for 1 k p j . We again remove these
points, and thus we have removed even number of points. And odd number of open
intervals is left, call the odd number 2q 1. Note that since the function f is monotonic in
every open interval left, R l ,1 l 2q 1, the image of f on these open interval is also an
open interval. If R a R b , sayR a a 1 , a 1 and R b b 1 , b 2 with (without loss of
generality) a 1 b 1 a 2 b 2 , then
R a R b by property A.
(Otherwise, b 1 is only point such that fx fb 1 , which contradicts property A.) Note
that given any R a , there must has one and only one R b such that R a R b . However, we
have 2q 1 open intervals is left, it is impossible. Hence, we know that f has infinite many
discontinuities on 0, 1.
4.28 In each case, give an example of a real-valued function f, continuous on S and
such that fS T, or else explain why there can be no such f :
(a) S 0, 1, T 0, 1.
Solution: Let
fx
2x if x 0, 1/2,
1ifx 1/2, 1.
(b) S 0, 1, T 0, 1 1, 2.
Solution: NO! Since a continuous functions sends a connected set to a connected set.
However, in this case, S is connected and T is not connected.
(c) S R 1 , T the set of rational numbers.
Solution: NO! Since a continuous functions sends a connected set to a connected set.
However, in this case, S is connected and T is not connected.
(d) S 0, 1 2, 3, T 0, 1.
Solution: Let
(e) S 0, 1 0, 1, T R 2 .
fx
0ifx 0, 1,
1ifx 2, 3.
Solution: NO! Since a continuous functions sends a compact set to a compact set.
However, in this case, S is compact and T is not compact.