Copy of s3 - Solutions to homework # 3. 1. For each...

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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.

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