1/25/2017
Solutions to Math 425 Homework 2
1. (Exercise 1.4.1)
(a) Suppose a, b ∈ Q, so a =
m
n
and b =
p
q
for some m, p ∈ Z and n, q ∈ N. Then
ab =
mp
nq
where mp ∈ Z and nq ∈ N, so ab ∈ Q. (Note that this fraction might not be shown in lowest terms.)
Similarly,
m p
mq + pn
a+b=
+ =
∈ Q.
n
q
nq
(b) Let a ∈ Q and t ∈ I, where I = {x ∈ R|x ∈
/ Q}. Then certainly a + t ∈ R and at ∈ R. Also, we
know (−a) ∈ Q.
Suppose for contradiction that a + t ∈ Q. Then (a + t) + (−a) ∈ Q, so t ∈ Q, a contradiction.
Therefore a + t ∈
/ Q. Since a + t ∈ R, we get a + t ∈ I.
Now assume a 6= 0, so that a1 ∈ Q. Suppose for contradiction that at ∈ Q. Then (at) a1 ∈ Q, so
t ∈ Q, a contradiction. Hence at ∈
/ Q. Since at ∈ R, this yields at ∈ I.
√
(c) Given s ∈ I and t ∈ I, it is possible that s + t ∈ I: for an example, take s = t = 22 . But it is also
√
√
possible that s + t ∈ Q: for example, take s = 2 and t = − 2.
√
Similarly, it is possible that st ∈√Q: for example, take s = t = 2. But it is also possible that
st ∈ I: for example, take s = t = 4 2.
So, given s, t ∈ I, the most we can really say for certain is that s + t ∈ R and st ∈ R.
T∞
2. (Exercise 1.4.3) For convenience, let S = n=1 (0, 1/n). We need to prove that there is no element x
such that x ∈ S.
Each interval (0, 1/n) is a subset of R, so the intersection of these intervals is a subset of R. Thus any
element of S would have to be an element of R, so it suffices to show that no element of R is an element
of S. Equivalently, we need to show that every element of R is not an element of S.
Let x ∈ R. If x ≤ 0 then x ∈
/ (0, 1/1), and S ⊆ (0, 1/1) so x ∈
/ S. So we may assume x > 0. By the
Archimedean Property (part ii) there exists n ∈ N such that 1/n < x, so x ∈
/ (0, 1/n). Thus x ∈
/ S, as
desired.
√
√
3. (Exercise
1.4.5)
√
√ Let a, b ∈ R, and suppose a < b. Then we know a − 2 ∈ R and b − 2 ∈ R, and
a − 2 < b − 2.
By Theorem 1.4.3 (the Density of Q in R), there exists r ∈ Q such that
√
√
a− 2<r <b− 2
Therefore
a<r+
√
2<b
√
√
But r ∈ Q and 2 ∈ I, so by Exercise 1.4.1, part (b), r + 2 ∈ I.
4. (Exercise 1.4.6)
1
(a) Not dense. For example, let a = 0 and b =
rational number pq with q ≤ 10. For, suppose
0<
1
1000 .
Between these two real numbers, there is no
1
p
<
q
1000
where p, q ∈ N. (Note that p and q must both be positive integers since pq > 0.) Then crossmultiplying yields 1000p < q, and does not switch the inequality since p, q > 0. But p ∈ N so p ≥ 1.
Therefore q > 1000p ≥ 1000(1) = 1000.
(b) This set is dense in R. One way to prove it would be similar to our proof of the Nested Interval
Property.
Let a, b ∈ R and suppose a < b. First, if there is an integer k such that a < k < b, then we’re done;
k = k1 = 2k0 . So we can assume that [a, b] ⊆ [k, k + 1] for some k ∈ Z. Let I1 = [k, k + 1].
Next, we define a closed interval I2 . We start by splitting I1 into two subintervals of equal length:
L1 = [k, k + 21 ] and R1 = [k + 12 , k + 1]. We see that L1 ∪ R1 = I1 , and L1 ∩ R1 = {k + 12 }. (These
two subintervals L1 and R1 are, respectively, the “left half” and the “right half” of I1 .) Our choice
for I2 will be whichever of these intervals contains a. If both subintervals contain a, then a = k + 21 ;
in that case choose I2 = R1 . Regardless of which interval we choose for I2 , note that the left and
right endpoints of I2 will be rational numbers whose denominator is a power of 2.
Next, we’ll split I2 in half to get L2 and R2 . The endpoints of I2 will be rational numbers, each
with a power of 2 for its denominator. Choose I3 to be R2 if a ∈ R2 , and L2 otherwise.
Continuing, we’ll split each closed interval In into two closed subintervals Ln and Rn of equal length,
whose union is In and whose intersection is a single point. Then In+1 is chosen to be Rn if a ∈ Rn ,
and In+1 = Ln otherwise.
Now for each n ∈ N, we have a ∈ In , and by construction a is not the right-hand endpoint of In .
1
Further, the length of In is 2n−1
, and each endpoint of In is a rational number with a power of 2
for its denominator.
1
Now, a < b so let = b − a > 0. There exists n ∈ N such that 2n−1
< . (This is equivalent to
1
(∃n ∈ N)(log2 ( ) < n − 1), which is true by the Archimedean Property.) Therefore the length of
In is less than the distance between a and b, so In can’t contain both a and b. We know that In
contains a, so it must not contain b. Finally, the right-hand endpoint of In is greater than a.
Let In = [g, h], so g ≤ a < h and b ∈
/ [g, h]. We can’t have b < g, since that would imply b < g ≤ a.
Therefore we must have b > h. Thus a < h < b, and h is a rational number whose denominator is
a power of 2.
(c) Not dense. Indeed, there is no such
p
q
between
1
1000
and
2
1000 .
For, suppose p ∈ Z and q ∈ N satisfy
1
p
2
< <
.
1000
q
1000
Then in particular, | pq | =
implies 10|p| < q.
p
q
<
2
1000 ,
so 1000|p| < 2|q| = 2q. Hence 500|p| < q, which certainly
5. (Exercise 1.4.8)
1
1
(a) Let A = {− 2n
|n ∈ N} and B = {− 2n+1
|n ∈ N}.
T∞
−1 1
(b) For each n ∈ N, let Jn = ( n , n ). Then n=1 Jn = {0}.
(c) For each n ∈ N, let Ln = [n, ∞).
(d) This is impossible. For each N ∈ N, let
JN = I1 ∩ I2 ∩ . . . ∩ IN =
N
\
In
n=1
Then each JN is a single closed interval, JN = [aN , bN ]. Proof: let In = [xn , yn ], so JN =
[x1 , y1 ] ∩ [x2 , y2 ] ∩ . . . ∩ [xN , yN ]. Then JN = {x ∈ R|xn ≤ x ≤ yn for all n ≤ N }. This set is equal
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to {x ∈ R| max(x1 , x2 , . . . xN ) ≤ x ≤ min(y1 , y2 , . . . yN )} = [max(x1 , x2 , . . . xN ), min(y1 , y2 , . . . yN )],
except that we must consider the possibility that both endpoints are the same–that JN consists of
a single point. If so, then JN = {m} for some m ∈ R. Either m ∈ Jk for all k, (in which case the
intersection of all the JM is nonempty) or there exists k such that m ∈
/ Jk (in which case k > N
and then Jk is empty). Either way, it’s a contradiction, so each JN is a closed interval of positive
length.
T∞
T∞
T∞
Now, notice that n=1 In = n=1 Jn . Proof: first suppose x ∈ n=1 In . Then for each
T∞n ∈ N, we
have x ∈ In . Thus for each NT∈ N, we have x ∈ I1 ∩ I2 ∩ . . . ∩ IN = JN . That is, x ∈ n=1 Jn . On
∞
the other hand, suppose x ∈ n=1 JnT
. Then for each n ∈ N, we have x ∈ I1 ∩ I2 ∩ . . . In ⊆ In ,so
∞
x ∈ In for each n ∈ N. Therefore x ∈ n=1 In .
Finally, we can see that each Jn+1 = Jn ∩ In+1 , so Jn ⊇ Jn+1 . Thus J1 ⊇ J2 ⊇ . . . is a chain of
nested closed intervals. The conclusion now follows from the Nested Interval Property.
6. (Exercise 1.5.2)
The intersection of the intervals must indeed have at least one element, but that element need not be
rational.
7. (Exercise 1.5.4)
(a) Assume a, b ∈ R and a < b. We know that (0, 1) ∼ R, so (by Exercise 1.5.5, part (c)) it would
suffice to show (a, b) ∼ (0, 1). Indeed, by Exercise 1.5.5, part (b) it would then suffice to show
(0, 1) ∼ (a, b).
Define f : (0, 1) → (a, b) by f (x) = a+(b−a)x. (Note that the domain and codomain are consistent:
if 0 < x < 1 then 0 < (b−a)x < b−a, since b−a > 0. Therefore a < a+(b−a)x < b, so a < f (x) < b.
Thus f really does send elements of (0, 1) to elements of (a, b).)
Show that f is one-to-one: Let x, y ∈ (0, 1) and suppose f (x) = f (y). That is, a + (b − a)x =
a + (b − a)y. Then (b − a)x = (b − a)y. Now since b − a > 0 we can divide by (b − a). The result is
x = y, as desired.
y−a
Show that f is onto: Let y ∈ (a, b). Then 0 < y − a < b − a, so y−a
b−a ∈ (0, 1). Now f ( b−a ) =
a + (b − a) y−a
b−a = a + (y − a) = y, so f is onto. (b) We know that the rationals in the open interval (0, 1) are countable, so we can arrange them in a
list, as Q ∩ (0, 1) = {t1 , t2 , t3 , . . .}. Now for each n ∈ N, let xn+1 = tn , and then let x1 = 0. Now
Q ∩ [0, 1) = {x1 , x2 , x3 , . . .}, and x1 = 0.
Now define g : [0, 1) → (0, 1) by
(
x
if x ∈
/Q
g(x) =
xn+1 if x = xn ∈ Q
Now, g is one-to-one: Let x, y ∈ [0, 1), and suppose x 6= y. If x and y are both irrational, then
g(x) = x and g(y) = y, so g(x) = x 6= y = g(y) and thus g(x) 6= y. If x and y are both
rational then x = xn and y = xm for some n, m ∈ N, and n 6= m. Then xn+1 6= xm+1 , so
g(x) = xn+1 6= xm+1 = g(y). Finally, if one of {x, y} is rational and the other is not, then one of
{g(x), g(y)} will be rational and the other not, so g(x) 6= g(y).
Finally, g is onto: Let y ∈ (0, 1). If y is irrational, then g(y) = y; notice that y ∈ (0, 1) ⊆ [0, 1) so y
is in the domain of g. On the other hand if y is rational, then y = xn for some n ∈ N, and n 6= 1
since y > 0 = x1 . So n − 1 ∈ N and thus g(xn−1 ) = xn−1+1 = xn = y. So g is onto.
8. (Exercise 1.5.5)
(a) Let A be a set, and define f : A → A by f (x) = x. Then f is one-to-one: suppose f (x) = f (y).
Then by definition of f , we have x = y. And f is onto; given y ∈ A, we have f (y) = y.
(b) Let A and B be sets, and suppose A ∼ B. That means there exists f : A → B such that f is
one-to-one and onto. Now define g : B → A by the following rule:
g(y) = the unique x such that f (x) = y
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To see that this definition makes sense, we need to show two things: (1) that for each y ∈ B there
is such an x ∈ A, and (2) that for each y ∈ B there is only one such x. Statement (1) is exactly the
statement that f is onto, which is true by hypothesis. Meanwhile statement (2) just asserts that f
is one-to-one: suppose y ∈ B and f (x1 ) = f (x2 ) = y. Then since f is one-to-one, it follows that
x1 = x2 .
Now that we know g is a correctly-defined function, let us show that g is one-to-one and onto.
Suppose g(y1 ) = g(y2 ) = x. Then f (x) = y1 , and f (x) = y2 . Therefore y1 = y2 .
Now suppose a ∈ A. Then f (a) ∈ B. We wish to show that g(f (a)) = a. Well, g(f (a)) =
the unique x such that f (x) = f (a). But clearly f (a) = f (a), so “the unique x ∈ A such that
f (x) = f (a)” must be a. Therefore g(f (a)) = a. (We have just shown that a function which is one-to-one and onto must have an inverse.)
(c) Suppose f : A → B and g : B → C are both one-to-one and onto. We wish to show that
(g ◦ f ) : A → C is also one-to-one and onto.
Suppose (g ◦ f )(x) = (g ◦ f )(y). That is, g(f (x)) = g(f (y)). Since g is one-to-one, we conclude that
f (x) = f (y). But since f is one-to-one, this yields x = y.
Now let z ∈ C. Since g : B → C is onto, there exists y ∈ B such that g(y) = z. Now since f is onto,
there exists x ∈ A such that f (x) = y. Thus (g ◦ f )(x) = g(f (x)) = g(y) = z, so (g ◦ f ) is onto.
9. (Exercise 1.5.7)
(Discussed in class)
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