The Real And Complex Number Systems

Proof: Since the hypothesis says that f is one-to-one, then by Theorem*, we know that f
is trictly monotonic on 0, 1.
Remark: (Theorem*) Under assumption of continuity on a compact interval, 1-1 is
equivalent to being strictly monotonic. We will prove it in Exercise 4.62.
4.27 Let f be a function defined on 0, 1 with the following property: 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.
(a) Prove that f cannot be continuous on 0, 1.
Proof: Assume that f is continuous on 0, 1, and thus consider max x0,1 fx and
min x0,1 fx. Then by hypothesis, there exist exactly two values a 1 a 2 0, 1 such
that fa 1 fa 2 max x0,1 fx, and there exist exactly two values b 1 b 2 0, 1
such that fb 1 fb 2 min x0,1 fx.
Claim that a 1 0anda 2 1. Suppose NOT, then there exists at least one belonging
to 0, 1. Without loss of generality, say a 1 0, 1. Sincef has maximum at a 1 0, 1
and a 2 0, 1, we can find three points p 1 , p 2 ,andp 3 such that
1. p 1 a 1 p 2 p 3 a 2 ,
2. fp 1 fa 1 , fp 2 fa 1 ,andfp 3 fa 2 .
Since fa 1 fa 2 , we choose a real number r so that
fp 1 r fa 1 r fq 1 ,whereq 1 p 1 , a 1 by continuity of f.
fp 2 r fa 1 r fq 2 ,whereq 2 a 1 , p 2 by continuity of f.
fp 3 r fa 2 r fq 3 ,whereq 3 p 3 , a 2 by continuity of f.
which contradicts the hypothesis that for every real number y, there are exactly two values
of x in 0, 1 for which fx y. Hence, we know that a 1 0anda 2 1. Similarly, we
also have b 1 0andb 2 1.
So, max x0,1 fx min x0,1 fx which implies that f is constant. It is impossible.
Hence, such f does not exist. That is, f is not continuous on 0, 1.
(b) Construct a function f which has the above property.
Proof: Consider 0, 1 Q c 0, 1 Q 0, 1, and write
Q 0, 1 x 1 , x 2 ,...,x n ,.... Define
1. fx 2n1 fx 2n n,
2. fx x if x 0, 1/2 Q c ,
3. fx 1 x if x 1/2, 1 Q c .
Then if x y, then it is clear that fx fy. Thatis,f is well-defined. And from
construction, we know that the function defined on 0, 1 with the following property: 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.
Remark: x : f is discontinuous at x 0, 1. Givena 0, 1. Note that since
fx N for all x Q 0, 1 and Q is dense in R, forany1ball Ba; r Q 0, 1,
there is always a rational number y Ba; r Q 0, 1 such that |fy fa| 1.
(c) Prove that any function with this property has infinite many discontinuities on
0, 1.
Proof: In order to make the proof clear, property A of f means that