Solutions to homework # 3.
1. For each positive integer
k
, there are only finitely many equations with integer coefficients
a
0
,
. . .
,
a
n
such that

a
0

+
· · ·
+

a
n

+
n
=
k
. Each such equation has
n
≤
k
roots. Let
A
k
denote all the roots of all equations with

a
0

+
· · ·
+

a
n

+
n
=
k
. Then the set of all
algebraic numbers is the union
∪
k
∈
IN
A
k
. This is a countable union of finite sets, therefore is
at most countable. The set of all algebraic numbers contains ZZ, hence is infinite, so it must
be countable.
2. The set IR of all real numbers is the (disjoint) union of the sets of all rational and irrational
numbers. We know from the lecture that IR is uncountable, whereas Q
is countable. If the
set of all irrational numbers were countable, then IR would be the union of two countable
sets, hence countable. Thus the set of all irrational numbers is uncountable.
3. Here is one of many examples of a bounded set with exactly three limit points:
S
:=
{
1
n
:
n
∈
IN
} ∪ {
1 +
1
n
:
n
∈
IN
} ∪ {
2 +
1
n
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 Spring '09
 Topology, Equations, Rational number, Countable set, Closed set, limit point

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