PROBLEM SET II
“DUST IN THE WIND”
DUE THURSDAY,  SEPTEMBER 
Definition. Define a function h : N × N × N . N in the following manner:

q+1
if n = 0;





if n = 1 and q = 0;
p
h(n, p, q) := 0
if n = 2 and q = 0;



1
if n ≥ 3 and q = 0;



h(n − 1, p, h(n, p, q − 1)) otherwise.
Let’s take a moment to unpack this. Let’s look at h(n, p, q) for fixed values of n.
Exercise . Show that we have the following equations for any p, q ∈ N.
h(0, p, q)
h(1, p, q)
h(2, p, q)
h(3, p, q)
=
=
=
=
q + 1;
p + q;
pq;
pq .
Let’s do some quick computations of some of the numbers h(n, p, q) to get a
feeling for the growth of this function.
Exercise . Compute explicitly the numbers h(4, 2, 4) and h(5, 3, 2). Show that
for any positive integer n, one has h(n, 2, 2) = 4.
Now let’s assemble this function of three variables into a function of a single
variable.
Definition. Define a function a : N .
N in the following manner
a(n) := h(n, n, n).
us
a(0) = 1,
a(1) = 2,
a(2) = 4
44
256
a(3) = 27, and a(4) = 44 = 44 .
e next value of this function, a(5), is a number so huge that it cannot be written
it down without new notation. is function (or, rather, one like it) is called the
Ackermann function.
Exercise . Show that the function a is injective.


DUE THURSDAY,  SEPTEMBER 
Exercise . For any natural number n, let Bn be the set of natural numbers m
such that a(m) ≤ n. Show that Bn is a finite set. Now for any natural number n,
we write
{
0
if Bn = ∅;
b(n) :=
max(Bn ) otherwise.
is defines a function b : N . N. Compute b(n) for 0 ≤ n ≤ 5000000. Despite this, show that b increases without bound; that is, show that for any natural
numbers r ≤ s, one has b(r) ≤ b(s), and for any natural number N, there exists a
natural number n such that b(n) > N.
Imagine making a table with the values of b. You’d see that the values were increasing unbelievably slowly. If you made an entry every nanosecond since the
beginning of time, you still wouldn’t have gotten to an n large enough to give you
b(n) = 4. You might even be tempted to conclude that the numbers would never
get above 3. Nevertheless, the previous exercise shows you that the numbers are
in fact increasing without bound.
Here is a very interesting subset of the real line. It has a variety of unusual features, which may come as a surprise to you. In particular, it may be the simplest
example of a actal.
Definition. e Cantor set C ⊂ R is the following set. Set C0 := [0, 1]. Proceed
iteratively in the following manner. For every natural number n > 0, set
Mn :=
3n−1
∪−1 (
k=0
3k + 1 3k + 2
,
3n
3n
)
Finally, set
C :=
and
∩
Cn := Cn−1 − Mn .
Cn .
n∈N
It is clear that C ⊂ [0, 1], so C is bounded.
Exercise . Draw Cn for 0 ≤ n ≤ 4. Show that C is closed, and that it contains
no nonempty open subset of R.
Exercise . Let Map(N+ , {0, 1}) denote the set of all maps N+ .
where N+ = {n ∈ N | n ≥ 1}. We aim to define a map
f: C.
Map(N+ , {0, 1}).
We define, recursively, a map
fn : Cn .
Map({1, 2, . . . , n}, {0, 1})
{0, 1},
PROBLEM SET II
“DUST IN THE WIND”

for every n ∈ N. For n = 0, there’s nothing to do, since there is exactly one map
∅ . {0, 1}. Now suppose the map fn−1 is defined; then set, for any point x ∈ Cn
and any integer 1 ≤ m ≤ n,


fn−1 (x)(m) if 1 ≤ m ≤ n − 1;
fn (x)(m) := 0
if (m = n) ∧ ∃k ∈ Z, x ∈ [3k/3n , (3k + 1)/3n ] ;

1
if (m = n) ∧ ∃k ∈ Z, x ∈ [(3k + 2)/3n , 3(k + 1)/3n ] .
Now we define the map f by the formula f (x)(m) := fn (x)(m) for any integer
n ≥ m. Show that f is a bijection.
Exercise⋆ . Prove that for any two distinct points x, y ∈ C, there exist disjoint
open subsets U, V ⊂ R such that x ∈ U, y ∈ V, and C ⊂ U ∪ V.