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MATH 355 HW # 2 SOLUTIONS
Here are your solutions to HW # 2, I’m still working on the grading but thought
you would enjoy these solutions. Enjoy!!!
2.2.3 There are three inequalities to prove, the middle inequality is obvious since
A is nonempty. So suppose A ⊆ B. We must prove inf(B) ≤ inf (A), and
sup(A) ≤ sup(B).
To prove the first inequality, we claim that inf(B) is a lower bound for A,
and we prove this first. Notice that by definition, inf(B) is a lower bound for
B, and as a result inf(B) ≤ b for every b ∈ B. But this is also the case for
every b ∈ A, since A ⊆ B. So, inf(B) is a lower bound for A. As a result,
the inf(A) must be greater than or equal to inf(B), since the inf(A) is the
greatest lower bound, and inf(B) is merely just one lower bound.
To prove the other inequality, we prove first that sup(B) is an upper bound
for A. By definition, sup(B) is an upper bound for B, so that sup(B) ≥ b for
every b ∈ B, and since A ⊆ B, it must be that sup(B) ≥ b for every b ∈ A.
The sup(A) is the least upper bound, which means it is less than or equal to
the upper bound sup(B).
2.2.9 Consider the sets A = B = {−1, 0}. Clearly sup(A) = sup(B) = 0. However,
by direct computation, C = {0, 1}, and so sup(C) = 1 6= sup(A) sup(B) = 0.
2.3.3 (b) sup(Q) = ∞, and inf(Q) = −∞.
(c) sup({n − n1 |n ∈ N}) = ∞, and inf({n − n1 |n ∈ N}) = 0.
2.3.4 Let x < y be given real numbers. Then by the density of Q in R, there exists
a rational number q with
√
√
x − 2 < q < y − 2.
It follows that
√
x < q + 2 < y,
√
and by Exercise 2.1.1, q + 2 is irrational.
2.4.9 We express A × B as the countable union of countable sets, and then deduce
that A × B is countable by Theorem 2.4. Notice that
A × B = ∪a∈A {(a, b)|b ∈ B}.
Each of the sets {(a, b)|b ∈ B} ∼ B, and since B is countable, so must
{(a, b)|b ∈ B} be countable. Further, the index set A is countable, and so
there are a countable number of these sets in this union. Therefore A × B is
the countable union of countable sets, and the result follows by Theorem 2.4.
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MATH 355 HW # 2 SOLUTIONS
2.4.10 Suppose to the contrary that A\B is countable, but that A is uncountable
and that B is a countable subset of A. Then we would have
A = (A\B) ∪ B,
and so A would be countable, as the union of the countable sets A\B and B.
But this contradicts the assumption that A is uncountable. (Note that this
is a proof that the set of irrational numbers is uncountable, where A = R,
and B = Q.)