Download Math 403 SOLUTIONS FOR HWK #4 RATIONAL, ALGEBRAIC

Math 403
SOLUTIONS FOR HWK #4
RATIONAL, ALGEBRAIC, IRRATIONAL AND TRANSCENDENTAL NUMBERS
Definition. A non-algebraic real number is called a transcendental number.
Denote the set of all irrational numbers by I, and the set of all transcendental
numbers by T. It is known that π and e are transcendental numbers.
It is obvious that R = Q ∪ I = A ∪ T. In HWK 3 you proved that Q ⊂ A.
PROBLEM A
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1) We know that 2 ∈ I. Does 2 ∈ A?
2) Prove that T ⊂ I (strict inclusion).
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2
√ Solution.1) Yes: 2 is a root of the polynomial x − 2 hencec 2 c∈ A hence
2∈
/ T 2) By Problem A in HWK #3, Q ⊂ A. Therefore T = A ⊂ Q = I. The
inclusion is strict by Part 1.
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PROBLEM B
1) What are the cardinalities of the sets T and I?
2) Prove that the set T is dense in R, i.e. that each interval (a, b) contains at
least one transcendental number. Does this imply that the set I is also dense?
Solution. 1) T = R \ A. Since A is countable, by Proposition 1.3.E, card(T) =
card(R) = c. Similarly, I = R \ Q. Since Q is countable, by Proposition 1.3.E,
card(I) = card(R) = c.
2) For each interval (a, b) the set T ∩ (a, b) = (a, b) \ (A ∩ (a, b)); since the set
A ∩ (a, b) is countable (as a subset of a countable set), card(T ∩ (a, b)) = card(a, b) =
c. Hence each interval contains transcendental numbers so T is dense. Since each
transcendental number is irrational, the set I is also dense (of course, this follows also
from the fact that card(I ∩ (a, b)) = card(a, b) = c
THE SET OF REALS R AS A METRIC SPACE
Recall that the absolute value |x| of a number x ∈ R is defined as |x| = x if x ≥ 0
and |x| = −x if x < 0. Of course, | − x| = |x|. In MATH 312, the following ”triangle
inequality” for the absolute value was proved: |a + b| ≤ |a| + |b|, a, b ∈ R.
Let ρ(x, y) = |x − y|, x, y ∈ R.
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PROBLEM C
Prove that ρ(x, y) ia metric in R.
Solution. Let us prove that the function ρ(x, y) possesses all properties of a
metric. For each x, y, z ∈ R,
a) ρ(x, y) = |x − y| ≥ 0.
b) ρ(x, y) = 0 ⇔ |x − y| = 0 ⇔ x − y = 0 ⇔ x = y.
c) ρ(x, y) = |x − y| = |y − x| = ρ(y, x).
d) ρ(x, y) = |x − y| = |(x − z) + (z − y)| ≤ |x − z| + |z − y| = ρ(x, z) + ρ(z, y) (by
the triangle inequality for the absolute value).
PROBLEM D
1) What is a ball B(a, r) in R?
2) Prove that each open interval (a, b), (a, ∞), (−∞, b) is an open set.
3) Is the interval (0, 1] open?
Solution. 1) B(a, r) = {x ∈ R : |x − a| < r} = {x ∈ R : −r < x − a < r} = {x ∈
R : a − r < x < a + r} = (a − r, a + r).
2) Let x ∈ (a, b). Put r = min(x − a, b − x). It is easy to see that (x − r, x + r) ⊂
(a, b). Thus each x ∈ (a, b) is an interior point of (a, b) thus each such interval is
open. If x ∈ (a, ∞) put r = x − a; again (x − r, x + r) ⊂ (a, ∞). Thus each x ∈ (a, ∞)
is an interior point of (a, ∞) thus each such interval is open. The intervals (−∞, b)
can be considered in a similar way.
3) No, the point 1 is not an interior point of this interval since for each ε > 0 the
ball (1 − ε, 1 + ε) is not contained in (0, 1].
An obvious but useful remark. Two open intervals are either disjoint or their
union is an interval. Moreover, you can use without proof the following intuitively
evident statement: If a collection of intervals contains a common point, their union
is an interval.
PROBLEM E
(3 pt)
Provide a straightforward proof of the statement:
a union of disjoint open intervals is an open set.
Solution. If x belongs to a union of intervals, it is contained in some interval
(a, b). According the statement in Problem D2, this interval is open so it contains a
ball (x − r, x + r) centered at x. Of course, this ball is also contained in the whole
union.
In the following problem you will prove that the converse statement is also true.
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PROBLEM F
Let U be an open set in (R, ρ). An open interval I ⊂ U is called a composite
interval in U if there is no other interval J ⊂ U such that I ⊂ J.
1) Prove that any two composite intervals are disjoint.
2) Let x ∈ U . Consider the union I of all intervals Ix such that x ∈ Ix and
Ix ⊂ U . Prove that x ∈ I and that I is a composite interval in U .
3) Prove that U is the union of its composite intervals.
Solution. 1) Assume that two composite intervals intersect. Then their union
is an interval which is contained in U and contains both component intervals. This
contradicts to the properties of a composite interval.
2) Let ∈ U . According to the above remark, the the union I of the intervals Ix
is an interval. x is contained in the intervals Ix , therefore it is contained in I. On
the other hand all Ix ⊂ U , therefore their union I ⊆ U . Now assume there is an
interval J such that I ⊂ J ⊆ U . It is clear that x ∈ J, and J is one of the intervals
Ix . Therefore it is contained in the union I. We came to a contradiction. This shows
that I is a composite interval.
3) Part 2 implies that each x in U belongs also to one of the composite intervals,
hence it is contained in their union Û . Hence U ⊆ Û . On the other hand, by
definition, each composite interval is contained in U , hence their union Û ⊆ U . Thus
U is the union of (disjoint) composite intervlas.
In Problems E and F you have proved a nice
Theorem. A set U ⊂ R is an open set iff it is a union of (disjoint) intervals.
We know that the collection of these intervals is countable!
Unlike in R2 , the open sets in R look very simple.
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