Targil 2.
(Questions about existence of weird analytical objects)

a
1. A {an} is a decreasing sequence of positive numbers,
n
n 1
is a divergent series
(‫)תור מתבדר‬. Consider the sequence {bn = min(an, 1/n)}.

Can we claim that
b
n 1
n
diverges?
Answer. Yes.
Solution. If we have finite number of bn = 1/n then except for the finite number of

elements bn = an and
 bn behaves the same way as
n 1

a
n 1
n
.
It remains to consider the case when for infinite number of indexes bn = 1/n.
Take the increasing sequence of all such indexes. Choose an infinite subsequence,
n1 , n2 , n3 ,... such that nk 1  2nk .

The series
 bn might be divided into sub-segments
n 1
nk

n  nk 1 1
bn . We shall show that

the sum in each sub-segment is at least ½, and that will imply
b
n 1
Indeed, for any n between nk-1 and nk, we have

n
 .

bn  min  an ,1 n   min ank ,1 nk  1 nk .
nk
Hence

n  nk 1 1
bn   nk  nk 1 
1 1
 . QED
nk 2
2.* Is it possible to construct f : 0,1  R which is continuous, monotone nondecreasing and satisfies the following two equations for every x  0,1 :
f  x 3  f  x  2
f  x   f 1  x   1
Answer: yes.
Solution. The idea comes from Cantor set. Cantor set is made as follows:
Take [0,1] interval. Exclude the open interval (1/3 , 2/3) in the middle, now you get
two intervals. Then exclude an open interval of length 1/9 in the middle of each
closed interval, you get 4 intervals. Repeat this operation infinite number of times
(excluding open interval in which is thrice shorter and located in the middle of
each remaining closed interval). What remains after infinite number of steps, is by
definition the Cantor set.
Another way to define Cantor set: take all numbers between 0 and 1, that can be
written by ternary fraction (base 3, unlike standard decimal or well-known to
computer programmers binary system) using only digits 0 and 2 (not using digit 1).
For instance 1 = 0.22222… base 3, ¼ = 0.0202020202…. base 3, etc.
Cantor set has many counter-intuitive properties. It is big in the sense of cardinality
(as big as the set of all real numbers) but small in the sense of “length” (set of
measure 0).
By the same idea we construct Cantor function, aka Cantor’s ladder. First we take
f (0) = 0 , f (1) = 1. On the interval [1/3, 2/3] we take f (x) = ½.
On each step of construction of Cantor set, we exclude an open interval in the
middle of a closed interval; by that moment, we have already defined values a and
b at the ends of that interval; the values on the closure of the open interval which is
being excluded will be (a + b)/2. This way we shall define the function at all points
of the Cantor set and at some points in it; the values at the rest of the points of the
Cantor set, such as ¼ , is defined by monotonicity.
Another way to define this function is as follows: for every x take the ternary
fraction representing x, which is 0.x1x2x3… (when representing 1, write 0.222…
and not 1.000…). Find the first appearance of digit 1 in the fraction (if it exists)
and erase both that digit and all digits coming after it. After that replace all the
digit 2 by digit 1 at all places; that will be the binary fraction of f (x).
After the example is constructed, it is not hard to verify that it satisfies the
condition; we leave it as an exercise to the reader.
3.* Derivative of a continuous function f : R  R at all rational points exists and
equals zero. Is f necessarily a constant?
Answer. No.
Solution. First, it is enough to do it for a closed interval. Indeed, we smoothly map
a real line into a closed interval, while rational numbers will go to rational
numbers, for instance as follows:
q : R   1,1
1
1  x2
If we build a function g on  1,1 satisfying all conditions, then function
q  x 
f (x) = g(q(x))
will satisfy all our conditions.
The set of rational numbers is countable. Therefore, we can assign a natural index
to each rational number.
We shall build some kind of Cantor’s ladder, but irregular (unlike the previous
Cantor’s ladder where we used precisely the middle 1/3 of each interval).
First define g  1  1, g 1  1.
Each time, when we have a closed interval, we shall choose on it a rational number
with the smallest index inside it; for that rational number we shall take an open
neighborhood inside that interval, with irrational ends. On the closure of this
neighborhood, the functions will be given a constant values, equal to the average of
the values at the ends of the closed interval. The missing values after infinite
number of steps are completed by monotonicity.
The Cantor function constructed in that way, for each rational number it is a
constant in a neighborhood, so it has a zero derivative at each rational number.
However, it isn’t constant.

4. {an} is a sequence of positive numbers,
a
n 1
n
 1 . Prove that it is possible to
insert into the open interval (0,1) a countable set of closed mutually disjoint
intervals, indexed by positive integers, so that a closed interval number n is of
length an.
Solution. We shall insert the intervals iteratively. First of all, we shall insert
interval of length a1, all the other intervals with odd indexes will be assigned to the
subinterval before a1, and all the intervals with even indexes will be assigned to the
subintervals after a1. The location of a1 interval will be chosen so that the length of
the before interval is equal to sum of lengths of all odd indexes > 1, and the length
of the after interval will be equal to sum of the lengths of all even indexes.
We shall continue the process inductively. On each stage, we have several open
intervals, to each an infinite set of closed intervals is assigned. From the closed
intervals assigned to each open interval, we take one the interval with the smallest
index, and separate the rest of them into two infinite subsets. The first subset is
assigned to the “before subinterval”, the second subset is assigned to the “after
subinterval”, the location of the chosen closed interval is chosen so that the lengths
if the before and the after subinterval will be precisely sufficient to cover the
intervals assigned to them.
At stage n, the interval of index n is already inserted, and all closed intervals are
disjoint; therefore this process will insert all intervals, and all will be disjoint.
5. f :  →  such that for any positive real x, y the sequence f (x+ny) , for n,
tends to infinity.
a. Can we claim that that f  x    ?
x 
b.** Can we claim that f  x    , if it is given that f is continuous?
x 
Answer. a. No. b. Yes
Solution. a. Let a > 1 be a transcendent number, and consider the following
function: f(an) = 0 for every natural n, and f(x) = x2 for all other points.
Any arithmetic progression can have no more than two common points with the
sequence an, since if it would have 3 common points, a would be a root of a
polynomial with rational coefficients. Hence any sequence f(x+ny) tends to
infinity, and f (x) doesn’t.
b. Suppose it doesn’t. Then for some M there is a sequence xk, converging to
infinity, such that f (xk) < M. Then, since f is continuous, for |x – xk| < εk , we have
f (xk) < 2M = N , where εk are small numbers, chosen separately for different k. So,
to get the contradiction we need to do one thing: build an arithmetic sequence
which intersects infinite subset of these small intervals.
Assume we have built a sequence {ny} which intersects K intervals:


nk y  xmk   mk , xmk   mk , for some indices nk , mk for k ≤ K.
We can move the y in certain interval so that the conditions


nk y  xmk   mk , xmk   mk still hold for the same nk , mk , because intersection of
open intervals is still an open interval, if it is nonempty.
We shall find such y that satisfies this condition for as large K as we want by
induction over K. For K = 1 it is obvious.


Assume that nk y  xmk   mk , xmk   mk , for given nk , mk for k ≤ K, in the interval
y  Y ,Y  , where α < 1.
Possible values of ny will cover the interval between ny and (n+1)y if
1 n 1 1

.
 , or n 
 n  1Y  nY i. e. 1  
n
n

1
So, all numbers above Y

will be covered by possible values of ny.
 1
But xk tends to infinity, so we can choose xm such that it can be equal to ny for a
certain value of y in the interval. This completes the induction.
By this inductive procedure we shall build an infinite set of indices mk and a nested
system of intervals  kYk , Yk  such that if we choose y in interval number K then


{ny} intersects intervals xmk   mk , xmk   mk for k ≤ K, the intersection of all those
intervals has at least on point y, and for that y sequence {ny} intersects infinite
number of intervals. Hence f (ny) doesn’t tend to infinity, contradiction, QED.