HOMEWORK 8
DUE NOVEMBER 7
(1) (a) Let a ∈ N. Show that a ≤ a(a+1)
. Deduce that there exists
2
(φ(a)−1)φ(a)
a unique φ(a) ∈ N such that
< a ≤ φ(a)(φ(a)+1)
,
2
2
defining a function φ : N → N.
Since a ∈ N, the required inequality is just 1 ≤ a+1
, by
2
canceling a on both sides. Since a ≥ 1, this is clear. Let
A = {b ∈ N|a ≤ b(b+1)
}. We have seen that a ∈ A, so
2
A 6= ∅. Thus, by induction, A has a minimal element,
which we call φ(a). Rest is clear.
(b) Show that any a ∈ N is equal to a = φ(a)(φ(a)+1)
− j where
2
0 ≤ j < φ(a). Thus, we get a new function ψ : N → N, by
ψ(a) = j.
By construction, since a ≤ φ(a)(φ(a)+1)
, we can write a =
2
φ(a)(φ(a)+1)
− j for some j ≥ 0. If j ≥ φ(a), then,
2
φ(a)(φ(a) + 1)
−j
2
φ(a)(φ(a) + 1)
≤
− φ(a)
2
(φ(a) − 1)φ(a)
=
2
< a.
a=
This is a contradiction.
(c) Define a function f : N → N × N, by f (a) = (φ(a) −
ψ(a), ψ(a) + 1). Show that this is a bijection.
As usual, we check that f is an injection and a surjection.
If f (a) = f (b), from the definition of f , it is clear that
φ(a) = φ(b) and ψ(a) = ψ(b). But, we know that,
φ(a)(φ(a) + 1)
− ψ(a)
2
φ(b)(φ(b) + 1)
=
− ψ(b)
2
= b.
a=
1
2
DUE NOVEMBER 7
To check, surjectivity, let (p, q) ∈ N×N. Let a = (p+q−1)(p+q)
−
2
q+1. I will leave you to verify that a ∈ N and f (a) = (p, q).
(2) If S ≈ Nn and x 6∈ S, show that S ∪ {x} ≈ Nn+1 .
By assumption, we have a bijection f : S → Nn . We define
a map F : S ∪ {x} → Nn+1 , by F (s) = f (s) if s ∈ S and
F (x) = n + 1. I will let you check that this is a bijection.
(3) Show that if S, T ⊂ Ω are finite sets (Ω need not be finite),
#(S ∪ T ) = #S + #T − #(S ∩ T ).
We induct on #T = n. If n = 0, T = ∅ and everything is
clear. If n = 1 (just to convince ourselves), T = {x}. If x 6∈ S,
this follows from the previous problem, since S ∩ T = ∅. If
x ∈ S, then S ∪ T = S, S ∩ T = {x} and the proof is clear.
So, assume result proved for n and we will prove this when
#T = n + 1. We first look at the case when T ⊂ S. Then
S ∪ T = S and S ∩ T = T and the result we need is clear. So,
assume that T is not a subset of S and let x ∈ T with x 6∈ S.
Denote by T 0 = T − {x}. Then, by the previous problem, we
have #T 0 = #T − 1. So, by induction, one gets,
#(S ∪ T 0 ) = #S + #T 0 − #(S ∩ T 0 ) = #S + #T − 1 + #(S ∩ T 0 ).
We have, by our choice, x 6∈ S∪T 0 and thus S∪T = (S∪T 0 )∪{x}
and by the previous problem, #(S ∪ T ) = #(S ∪ T 0 ) + 1, which
gives, #(S ∪ T ) = #S + #T − #(S ∩ T 0 ). Since x 6∈ S, we also
have S ∩ T 0 = S ∩ T , which gives us what we want.
(4) If S ⊂ Nn , show that S is finite (by induction, for example) and
#S ≤ n.
We prove this by induction. If n = 1, S ⊂ {1} = N1 and so
either S = ∅ or S = {1} = N1 . So, S is finite and #S ≤ 1.
So, assume the result proved for n and we will prove this
for n + 1. So, S ⊂ Nn+1 . If n + 1 6∈ S, then S ⊂ Nn and
thus by induction, S is finite and #S ≤ n < n + 1, proving
what we want. So, assume n + 1 ∈ S. If S = Nn+1 , again the
result is clear, so we may also assume that S 6= Nn+1 . So, let
k ∈ Nn+1 , k 6∈ S. Then, as we did in class, define a bijection
θ : Nn+1 → Nn+1 , by θ(i) = i if i 6= k, i 6= n + 1, θ(k) =
n + 1, θ(n + 1) = k. Under this map, we get a bijection θ : S →
S 0 ⊂ Nn+1 and θ(k) = n + 1 6∈ S 0 . Thus, S 0 ⊂ Nn and so by
induction, S 0 and thus S are finite and #S = #S 0 ≤ n < n + 1.
(5) If f : S → T is a surjective map with S finite, show that there
exists a subset S 0 ⊂ T (This was a typo. It should have been
S 0 ⊂ S) such that f|S 0 : S 0 → T is a bijection. Deduce that T
is a finite set and #T ≤ #S.
HOMEWORK 8
3
We prove this induction on #S. If S = ∅ or #S = 1, this is
obvious. So, assume that we have proved this for all finite sets S
with #S ≤ n and we will prove this when #S = n+1. If T = ∅,
no such function can exist and thus we may assume T 6= ∅. Let
t ∈ T and let S 0 = S − f −1 (t). Then S 0 ⊂ S and f −1 (T ) 6= ∅,
since f is surjective. The restriction of f : S 0 → T − {t} is
surjective (why?) and since S 0 is a finite set with #S 0 < #S,
we see that T − {t} is a finite set and #(T − {t}) ≤ #S 0 , by
induction. But, then T is a finite set (by an earlier problem)
and #T = #(T − {t}) + 1 ≤ #S 0 + 1 ≤ S, since f −1 (t) 6= ∅.
(6) Let S be a finite set with #S = n.
(a) Show that P(S) is a finite set and its cardinality is 2n .
We prove this by induction on n. If n = 1, this is clear.
So, assume result proved for n and we will prove this for
n + 1. So, let S be a set with #S = n + 1 and fix an
element s ∈ S. Then, we consider X = {A ⊂ S|s 6∈ A}
and Y = {A ⊂ S|s ∈ A}. We clearly have P(S) = X ∪ Y
and X ∩ Y = ∅. It is also clear that X = P(S − {s})
and so by induction, since #(S − {p}) = n, we have X is
finite and #X = 2n . We have a map φ : X → Y , given
by φ(A) = A ∪ {s}. I claim that this map is a bijection.
If φ(A) = φ(B), we have A ∪ {s} = B ∪ {s} and thus
A ⊂ B ∪ {s}. Since s 6∈ A, we see that A ⊂ B and
similar argument shows B ⊂ A and thus A = B, proving
injectivity. If B ∈ Y , clearly A = B − {s} ∈ X and
φ(A) = B, proving surjectivity. Thus Y too is finite and
#Y = #X = 2n . Now, from an earlier problem, we have,
#P(S) = #X + #Y = 2n + 2n = 2n+1 .
Thus, by induction, we have proved the result.
(b) Let r be an integer with 0 ≤ r ≤ n and let Cr (S) ⊂ P(S)
be the set of subsets A of S with #A = r. Show that
#Cr (S) = nr .
The proof is very similar to the previous problem. We use
induction on n and I will let you check the cases n = 0, 1.
So, assume that we have proved the result for n and we
prove it for n+1 and so let S be a finite set with #S = n+1.
If r = 0,we have only one such set, namely ∅ and since
1 = n+1
, we are done. Similarly, if r = n + 1, we have
0
only one such subset, namely all of S and since 1 = n+1
,
n+1
we are done. So, assume 0 < r < n + 1 and fix an s ∈ S.
Let X, Y be defined as in the previous problem and let
4
DUE NOVEMBER 7
X 0 = X ∩ Cr (S), Y 0 = Y ∩ Cr (S). Then X 0 = Cr (S − {s})
and thus by induction,#X 0 = nr . We also have a map
φ : Cr−1 (S − {s} → Y 0 , given by φ(A) = A ∪ {s}. As
before, one checks that φ is a bijection and thus #Y 0 =
#Cr−1
which again by induction, we know to be
(S − {s}),
n
0
. Since X ∩ Y 0 = ∅ and X 0 ∪ Y 0 = Cr (S), we have
r−1
n
n
n+1
0
0
#Cr (S) = #X + #Y =
+
=
,
r
r−1
r
the last equality from a previous set of home work. This
finishes the proof. P
(c) Deduce the formula nr=0 nr = 2n .
This is clear, since P(S) = ∪nr=0 Cr (S) and this union is
pairwise disjoint. So,
n n
X
X
n
n
#Cr =
.
2 = #P(S) =
r
r=0
r=0
(7) If A, B are finite sets with #A = m, #B = n at least one of m, n
non-zero, show that F (A, B) is a finite set and its cardinality
is nm .
Induct on m. You can also induct on m, but it is messier. If
you can not solve this exercise, may be we should have a talk.