SOLUTIONS FOR HOMEWORK 2
1.3.8.
For
k
∈
N
let
A
k
=
{
1
,
2
,...,k
}
. Then

A
k

=
k
, yet
∪
∞
k
=1
A
k
=
N
.
1.3.12.
Let
F
m
be the family of subsets of
N
with precisely
m
elements.
Let
N
m
=
N
×
...
×
N
(
m
times) be the
m
fold Cartesian product of
N
with itself,
that is, the set of all ordered
m
tuples whose elements are positive integers.
Use
induction on
m
to prove that
N
m
is countable.
The case of
m
= 1 is clear.
The
case of
m
= 2 was proved in class: there exists a bijection from
N
to
N
×
N
. This
forms the base of induction.
Now, suppose
N
m
is countable, and prove the same
for
N
m
+1
.
By the induction hypothesis, there exists an injection
f
m
:
N
m
→
N
.
For
a
= (
a
1
,...,a
m
,a
m
+1
)
∈
N
m
+1
, let
a
′
= (
a
1
,...,a
m
).
Define
f
m
+1
by setting
f
m
+1
(
a
) =
f
2
(
f
m
(
a
′
,a
m
+1
)).
You can check that
f
m
+1
is injective, hence
N
m
is
countable.
Next we define a map
g
:
F
m
→
N
m
, taking a set of
m
elements to the
m
tuple, listing
them in the increasing order. This is an injection, hence so is
f
m
+1
◦
g
. Therefore,
F
m
is countable.
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 Spring '08
 JUNGE
 Sets, Natural number, Countable set, countable union, countable sets.

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