MATH 203A section 33 Analysis on Rn (I)
Problem session 1
Tuesday, Oct. 7
Note: Though not all exercises in Rudin are assigned, you are encouraged to solve all
of them.
1. Rudin Ex.1: Proof by contradiction, using the fact that Q is a field.
2. Rudin Ex. 2: First argue that the problem can be reduced to show that there is no
rational number whose square is 3. Then argue like in class, say proof by contradiction:
assume r = p/q, p, q ∈ Z, q > 0, the greatest common divisor of p and q is one (that
is, q and p are relatively prime), and assume r 2 = 3. Derive a contradiction.
3. Rudin Ex. 3: Very basic.
4. Rudin Ex. 4: Very basic.
5. Rudin Ex. 5: Use definitions. Homework.
6. Assume existence (of R) theorem. Show that for any x ∈ R,
x = sup{q ∈ Q | q < x}.
Proof. If we study the construction of R as in Appendix of Rudin, this is implied by
definition. Here we proceed without knowing the construction.
To show x = sup A where A = {q | q ∈ Q, q < x}, we need to verify:
(1) x is an upper bound of A. Let us mention that A is not empty. Indeed, x − 1 ∈ R
and x − 1 < x. By the fact that Q is dense in R, there exists q ∈ Q such that
x − 1 < q < x.
In particular, it shows us that q ∈ A and hence A 6= ∅. By definition, any a ∈ A
satisfies a < x and hence a ≤ x so x is an upper bound of A.
(2) x is the least upper bound. For any ǫ > 0 we have x − ǫ < x. By density of Q we
know there exists q ∈ Q such that
x − ǫ < q < x.
Therefore, we have q ∈ A and x − ǫ < q and hence the conclusion.
7. Assume the existence theorem. For any x, y ∈ R, set
C = {t | t ∈ Q, t < x + y},
D = {r + s | r, s ∈ Q, r < x, s < y}.
Show that C = D.
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Proof. Strategy: we need to show that C ⊂ D and D ⊂ C.
(1) Firstly, D ⊂ C is easy. Indeed, take an arbitray element d ∈ D, it can be written
as d = r + s for some r, s ∈ Q, r ≤ x and s ≤ y. Then d = r + s < x + y and d ∈ Q
by the fact that Q is a field. Hence d ∈ C. By arbitrariness of d, we proved D ⊂ C.
(2) To show C ⊂ D, consider an arbitrary t ∈ C, that is t ∈ Q and t < x + y. Then
we have t − x < y. By (again!) density there exists q ∈ Q such that
t − x < q < y.
Set r = t − q and s = q then r, s ∈ Q, r < x, s < y and
t = r + s.
This shows t ∈ D and hence C ⊂ D.
8. Suppose A ⊂ R and B ⊂ R are two non-empty subsets bounded from above and
A ⊂ B. Prove that sup A ≤ sup B.
Proof. By the setting, α = sup A and β = sup B are two real numbers. For any a ∈ A
we have a ∈ B also and hence a < β. This shows β is an upper bound for A. Since α
is the least upper bound of A, we have α ≤ β.
9. Rudin Ex. 6. To make things more clear, let us assume that for the given b > 1. We
set
n
1
0
−n
b = 1,
and for n > 0, b =
.
b
(0) We easily check that bm > 0 for all m ∈ Z.
Proof. This is easy using the definitions above.
(1) We check that bm+n = bm bn for m, n ∈ Z.
Proof. This is easy using the definitions above.
(2) We check that m < n, m, n ∈ Z, bm < bn . In particular, for m ∈ Z, bm > 1 if
m > 0 and bm < 1 if m < 0.
Proof. This can be checked easily, say by considering the case when n ≤ 0, the
case when m < 0 < n and the case when m ≥ 0.
(3) For r ∈ Q, it can be written as r = p/q where p, q ∈ Z and q > 0. We will define
br = (bp )1/q . Note this quantity is well defined because bp > 0 and q is a positive
integer. To show br such defined does not rely on the representation of r as ratios
of two integers, we should check that if r = p/q = m/n where p, q, m, n ∈ Z and
q, n > 0, then
(bm )1/n = (bp )1/q .
Note that br > 0 for any r ∈ Q.
Proof. This can be proved by showing that both numbers, when raised to nq (which
is positive integer) power, yield the same number bmq = bnp . By uniqueness of
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(bmq ) nq , these two numbers are equal.
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(4) Prove br+s = br bs when r, s ∈ Q.
Proof. Since r, s, r + s ∈ Q, we have, for some m, n, p, q ∈ Q, n > 0, q > 0
r=
m
mq
=
,
n
nq
s=
p
pn
=
,
q
qn
r+s=
mq + np
.
nq
Then left hand side in (4) is
b
mq+np
nq
,
and its nq (which is a positive integer) power is bmq+np . The right hand side of
(4) is
mq pn
b nq b qn
and its nq power is bmq bpn = bmq+pn (say by step (1)). Hence again, by the
uniqueness of nq root, the two sides of (4) agree.
(5) Show that for any r ∈ Q, show that br > 1 if r > 0 and br < 1 if r < 0.
Proof. Let r = m/n, m, n ∈ Z and n > 0. If r > 0, then m > 0 and by definition
(br )n = bm > 1 and hence br > 1 also because otherwise (br )n ≤ 1 which is not
true. Since br b−r = 1, we verify that br < 1 if r < 0 since −r > 0 implies
b−r > 1.
(6) Show that for any r, s ∈ Q, r < s, it holds that br < bs .
Proof. In this case s = r + t for some t ∈ Q and t > 0. By (4) we have bs = br bt .
By (5) we have bt > 1. Since br > 0 and R is an ordered field we have bt br > br ,
that is bs > br .
(7) Given x ∈ R. Define B(x) = {bt | t ∈ Q, t ≤ x}. Prove that if r ∈ Q,
br = sup B(r).
Proof. By the monotonicity proved in (6), we have for that br > bt for any t ∈ Q
and t < r. Hence br is an upper bound for B(r). Meanwhile, br ∈ B(r). Hence
br = sup B(r) = max B(r). That is, the supremum is in fact achived when r ∈ Q.
(8) Given x ∈ R. If instead we define B ′ (x) = {bt | t ∈ Q, t < x}. Prove that
sup B(x) = sup B ′ (x).
Proof. If x 6∈ Q, then B(x) = B ′ (x) and hence sup B(x) = sup B ′ (x). Hence we
assume x = r ∈ Q below.
Since r ∈ Q, sup B(r) = br and for all t < x, by monotonicity in step (6) we
have bt < br . Hence br is an upper bound for B ′ (r). To show it is the least upper
bound, consider any real number c < br , we may assume c > 0 since other wise
it cannot be an upper bound because bs > 0 for all s ∈ Q. Set t = br /c > 1. By
Exercise 11(4) below (you proved this in HW1), there is some big n ∈ Z such
that b1/n < t. Then b−1/n > 1/t, and we have
b−1/n br > (1/t)br = c.
That is we have br−1/n > c. Note r − 1/n ∈ Q and r − 1/n < r. So we find
br−1/n ∈ B ′ (r) and c < br−1/n . So any c < br cannot be an upper bound of B ′ (x).
This proves that br = sup B ′ (r). So sup B ′ (r) = sup B(r) holds true for r ∈ Q
also.
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(9) Prove that bx+y = bx by . Hint: This is a bit easier if we use definition (8).
Proof. The proof is simpler if we use the definition of bx in step (8). This is
because of the fact we proved in Exercise 7 above, and the strict < there cannot
be replaced by ≤ as seen in Exercise 10 below.
Our goal is to show that bx by is the supremum of B ′ (x + y). Exercise 7 tells us
{t | t ∈ Q, t < x + y} = {r + s | r, s ∈ Q, r < x, s < y}.
This implies
{bt | t ∈ Q, t < x + y} = {br+s | r, s ∈ Q, r < x, s < y}.
Hence for any bt ∈ B ′ (x + y), we have bt = br+s = br bs for some r, s ∈ Q and
r < x, s < y. In particular, r ∈ B ′ (x) and s ∈ B ′ (y) and hence br ≤ bx , bs ≤ by .
This yields bx+y = sup B ′ (x + y) ≤ bx by .
Secondly, for any c < bx by , we set u = c/bx < by , v = 21 (u + by ) < by and
ρ = u/v. Note that ρ < 1, and we set d = bx ρ < bx .
Since by = sup B ′ (y) and v < by , there exists bs with s ∈ Q and s < y such
that bs > v. Similarly, since bx = sup B ′ (x) and d < bx , there exists br with
r ∈ Q and r < x such that br > d. Finally, r + s ∈ Q and r + s > x + y
and br+s > dv = bx u = c. Hence c cannot be an upper bound of B ′ (x + y) and
bx by = sup B ′ (x + y).
10. Let x, y ∈ R and define
C = {t | t ∈ Q, t ≤ x + y},
D = {r + s | r, s ∈ Q, r ≤ x, s ≤ t}.
Show that D ⊂ C but C is not always a subset of D. (Compare this with Problem
7!!!)
Proof. D ⊂ C can be proved as before.
Next, √
we give an example such that C is not a subset of D. Consider x =
y = − 2. Then
√
√
C = {t | t ∈ Q, t ≤ 0},
D = {r + s | r, s ∈ Q, r ≤ 2, s ≤ − 2}.
√
2 and
Clearly, 0 ∈ C. We√claim that 0 6∈√D. Indeed,√suppose 0 ∈ D then 0 = √
r + s for some
2 and s ≤ − 2. Since 2 is not
r, s ∈ Q and r ≤ √
√ rational, r < 2 must hold.
Hence s = −r > − 2 which contradicts with s ≤ − 2. So 0 ∈ C but 0 6∈ D and
hence C is not a subset of D.
11. Given 0 < a < b, n ∈ N and n ≥ 1.
(1) Prove that bn − an ≥ n(b − a)an−1 . Prove that bn − an ≤ n(b − a)bn−1 .
Proof. Recall the identity (which can be verified directly)
bn − an = (b − a)(bn−1 + bn−2 a + bn−3 a2 + · · · + ban−2 + an−1 ).
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If we replace b by a in the second parenthesis on the right, the value on the right
will be decreased, hence, noticing that there are totally n terms there
bn − an ≥ (b − a)(an−1 + an−2 a + · · · + aan−2 + an−1 ) = n(b − a)an−1 .
If we replace a by b in the second parenthesis on the right, the value on the right
will be increased, hence, noticing that there are totally n terms there
bn − an ≥ (b − a)(bn−1 + bn−2 b + · · · + bbn−2 + bn−1 ) = n(b − a)bn−1 .
(2) In particular, for any positive integer n. bn − 1 ≥ n(b − 1).
Proof. Set a = 1 above.
(3) Plug b1/n to the inequality above (note that b1/n > 1 and hence the inequality
can be applied), we have b − 1 ≥ n(b1/n − 1).
(4) If t > 1 and the integer n satisfies n > (b − 1)/(t − 1), then b1/n < t.
Proof. Use (3) and the lower bound of n, we have
b1/n ≤
b−1
b−1
+1<
+ 1 = t.
n
(b − 1)/(t − 1)
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