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Unformatted text preview: Solutions to homework # 3.
1. For each positive integer k, there are only ﬁnitely many equations with integer coeﬃcients
a0 , . . ., an such that a0  + · · · + an  + n = k. Each such equation has n ≤ k roots. Let
Ak denote all the roots of all equations with a0  + · · · + an  + n = k. Then the set of all
algebraic numbers is the union ∪k∈IN Ak . This is a countable union of ﬁnite sets, therefore is
at most countable. The set of all algebraic numbers contains Z hence is inﬁnite, so it must
Z,
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
1
1
: n ∈ IN} ∪ {1 + : n ∈ IN} ∪ {2 + : n ∈ IN}.
n
n
n The set lies within the interval [0, 3], therefore is bounded. The points 0, 1, 2 are the only
limit points. Indeed, a limit point of S cannot be smaller than 0 or bigger than 3 because of
the above bounds on S. All points in (0, 1) are either isolated points of S or not in S and
have a small neighborhood that does not intersect S. Likewise for all points in (1, 2) and in
(2, 3].
4. Yes, every point of any open set E in IR2 is a limit point of E, since every neighborhood
of a point in IR2 contains inﬁnitely many points. Since any point in E has a neighborhood
consisting of points of E only, any smaller neighborhood consists entirely of points from E
as well, so every point of E is its limit point.
The same is not true for closed sets in IR2 . For example, any ﬁnite nonempty set in IR2 is
closed but none of its points is its limit point. ...
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This document was uploaded on 11/17/2010.
 Spring '09
 Equations

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