Exercise of Chapter 2
Written by Hsin-Jung, Wu.
My mail is [email protected]
1. Discuss whether the following sets are open or closed.
(a) ]1,2[ in R1 = R
(b) [2, 3] in R
(c)
∞
T
∈ 1, 1/n[ in R
n=1
(d) Rn in Rn
(e) A hyperplane in Rn
(f) {r ∈]0, 1[| r is rational} in R
(g) {(x, y) ∈ R2 | 0 < x ≤ 1} in R2
(h) {x ∈ Rn | kxk = 1} in Rn
Solution: It is easy, then we omit it.
2. Determine the interiors, closures, and boundaries of the sets in Exercise 1.
Solution: It is easy, then we omit it.
3. Let U be open in M and U ⊂ A. Show that U ⊂ int(A). What is the corresponding statement for closed sets?
Solution: The proof you can use the definition to prove.
Let V be closed in M and A ⊆ V , then cl(A) ⊆ V .
4. (a) Show that if xn → x in a metric space M , then x ∈ cl{x1 , x2 , . . . }. When
is x an accumulation point?
1
Solution: It is easy to check x is an accumulation point of {x1 , x2 , . . . },
then we omit it. If x ∈
/ {x1 , x2 , . . . }, and x is a cluster point of {x1 , x2 , . . . },
then x must be an accumulation point.
(b) Can a sequence have more than one accumulation point?
Solution: Consider the sequence
 1
 k
xn =
 k+1
k
if n = 2k − 1
if n = 2k
then 1 and 0 are accumulation points of this sequence.
(c) If x is an accumulation point of a set A, prove that there is a sequence of
distinct points of A converging to x.
Solution: Given ε1 = 1 > 0 , there is a point x1 ∈ A such that 0 <
d(x, x1 ) < 1. Let ε2 = d(x, x1 ), there is a point x2 such that 0 < d(x, x2 ) <
d(x, x1 ) = ε2 . Let ε3 = d(x, x2 ), repeat the progress, we will have a
sequence xn and xi 6= xj if i 6= j, and xn converges to x.
5. Show that x ∈ int(A) iff there is an ε > 0 such that D(x, ε) ⊂ A.
Solution: x ∈ int(A), there is an open set U contains x, such that U ⊆ A. Since
U is open, there is an ε > 0, such that D(x, ε) ⊆ U , then D(x, ε) ⊆ A. On the
other side, since x ∈ D(x, ε) ⊆ A, and D(x, ε) is open, then x ∈ int(A).
6. Find the limits, if they exist, of these sequences in R2 :
(a)
(b)
(c)
(d)
(−1)n , n1
1
n
1,
1
n
(cos nπ),
1
n
sin nπ +
π
2
1
, n−n
n
Solution: It is easy to calculate, then I omit it.
2
7. Let U be open in metric space M . Show that U =cl(U )\bd(U ). Is this true for
every set in M .
Solution: Suppose x ∈ U , then x ∈ cl(U ), since U is open, there exists an
ε > 0, such that D(x, ε) ⊆ U , so x ∈
/ bd(U ). Hence U ⊆ cl(U )\bd(U ). Suppose
x ∈cl(U )\bd(U ), if x ∈ U , then done. Otherwise x is an accumulation point
of U . Then for each ε > 0, there is a y ∈ D(x, ε), where y ∈ U and we know
x ∈ U c , then x ∈ bd(A). Hence we get a contradiction, so cl(U )\bd(U ) ⊆ U .
No, choose U = {0}, cl(U ) = U , and bd(U ) = U . But cl(U )\bd(U ) = ∅.
8. Let S ⊂ R be nonempty, bounded below, and closed. Show that inf(S) ∈ S.
Solution: Claim inf(S) is an accumulation point of S. Let inf(S) = s. For
all ε > 0, there is x ∈ S such that s − ε < x < s + ε. So we have s is an
accumulation point of S, and S is closed, then s ∈ S.
9. Show that
(a) intB = B\ bdB, and
Solution: Suppose x ∈ int(B), there exists an ε > 0, such that D(x, ε) ⊆ B,
so x ∈bd(B).
/
Then int(B) ⊆ B\bd(B).
Suppose x ∈ B\bd(B), there exists an ε > 0, such that D(x, ε) ⊆ B, so
x ∈int(B). Then B\bd(B) ⊆ int(B).
(b) cl(A) = M \int(M \A)
Solution: Suppose x ∈ cl(A). If x ∈ A, then x ∈ M \(M \A). So x ∈
M \int(M \A). If x is an accumulation point of A, then for all ε > 0, there
is a y ∈ D(x, ε), where y ∈ A. So there is no D(x, ε) ⊆ M \A, thus x ∈
/
int(M \A).
Suppose x ∈
/ int(M \A), for all ε > 0, there is no D(x, ε) ⊆ M \A. Then
there is a y ∈ D(x, ε), where y ∈ A. So x is an accumulation point of A if
x 6= y, and x ∈ A if x = y. Thus x ∈ cl(A).
3
10. Determine which of the following statements are true.
(a) int(cl(A)) =int(A).
Solution: False. Let A = Q, int(cl(Q)) = int(R) = R. But int(Q) = ∅.
(b) cl(A)
T
A = A.
Solution: True. It is trivial to see cl(A)
T
A ⊆ cl(A) A.
T
A ⊆ A. But A ⊆ cl(A), then
(c) cl(int(A)) = A.
Solution: False. Let A = (0, 1), cl(int(A)) = [0, 1] 6= (0, 1) = A.
(d) bd(cl(A)) = bd(A)
Solution: False. Let A = Q, bd(cl(A)) = bd(R) = ∅, but bd(Q) = R.
(e) If A is open, then bd(A) ∈ M \A.
Solution: True. A is open implies A = int(A). For all x ∈ bd(A), for all
ε > 0 there is a y ∈
/ A , so y ∈ D(x, ε) then x is not in int(A)\A
11. Show that in a metric space xm → x iff for every ε > 0 there is an N such
that m ≥ N implies d(xm , x) ≤ ε (this differs from Position 2.7.2 in that here
” < ε” replaced by ” ≤ ε”).
Solution: One side is trivial. The other side is that for every ε > 0, there is a
positive integer N such that m ≥ N implies d(xm , x) ≤
ε
2
< ε. So xm → x. 12. Prove the following properties for subsets A and B of a metric space:
(a) int(int(A)) = int(A).
Solution: Since int(A) is an open set, and we know if a set S is open, then
int(S) = S. Hence int(int(A)) = int(A).
4
(b) int(A
S
B) ⊃ int(A)
S
(c) int(A
T
B) = int(A)
T
int(B).
S
Solution: For all x ∈ int(A) int(B), may assume x ∈ int(A). There
S
S
exists an ε > 0, such that D(x, ε) ∈ A ⊆ A B, hence x ∈ int(A B).
int(B).
T
Solution: For all x ∈ int(A B)
⇐⇒ there exist an ε > 0, D(x, ε) ∈ A
T
B
⇐⇒ there exist an ε > 0, D(x, ε) ∈ A and D(x, ε) ∈ B
⇐⇒ x ∈ int(A) and x ∈ int(B)
T
⇐⇒ x ∈ int(A) int(B)
13. Show that cl(A) = A
S
bd(A).
Solution: Suppose x is an accumulation point of A and x ∈ Ac , then for each
ε > 0, there exists y ∈ A such that y ∈ D(x, ε) and we know x ∈ Ac , then x ∈
S
bd(A). So cl(A) ⊆ A bd(A).
Suppose x ∈ bd(A)\A, for each ε > 0, there exists y ∈ A such that y ∈ D(x, ε),
S
then x is an accumulation point of A. Hence A bd(A) ⊆ cl(A).
14. Prove the following for subsets of a metric space M:
(a) cl(cl(A)) = cl(A).
Solution: Since cl(A) is an closed set, then cl(cl(A) = cl(A).
(b) cl(A
S
B) = cl(A)
S
(c) cl(A
T
B) ⊂ cl(A)
T
cl(B).
S
S
Solution: Since A ⊆ A B, then cl(A) ⊆ cl(A B). Similarly cl(B) ⊆
S
S
S
cl(A B). So cl(A) cl(B) ⊆ cl(A B).
S
S
S
S
Since A B ⊆ cl(A) cl(B) and cl(A) cl(B) is closed, then cl(A B) ⊆
S
cl(A) cl(B).
cl(B).
5
T
T
T
Solution: Since A B ⊆ A, A B ⊆ B, then cl(A B) ⊆ cl(A) and
T
T
T
cl(A B) ⊆ cl(B). So cl(A B) ⊆ cl(A) cl(B).
15. Prove the following for subsets of a metric space M:
(a) bd(A) =bd(M \A).
Solution: For all x ∈ bd(A), choose ε > 0, there is a y ∈ A and z ∈ M \A
such that y and z both ∈ D(x, ε). Then x ∈ bd(M \A).
Conversely is the same.
(b) bd(bd(A)) ⊂ bd(A).
Solution: For all x ∈ bd(bd(A)), choose
ε
2
> 0, there is a y ∈ bd(A) such
that y ∈ D(x, 2ε ). Since y ∈ bd(A), choose
ε
2
> 0, there is a y 0 ∈ A and
z 0 ∈ M \A such that y 0 and z 0 ∈ D(y, 2ε ) ⊆ D(x, ε), then x ∈ bd(A).
S
S S
bd(B) ⊂ bd(A B) A B.
S
S
Solution: For all x ∈ bd(A B), choose ε > 0, there is a y ∈ A B and
S
z ∈ M \(A B) such that y and z ∈ D(x.ε). W.L.O.G. let y ∈ A, and
S
z ∈ M \A. Then x ∈ bd(A) ⊆ bd(A) bd(B).
S
S
For all x ∈ bd(A) bd(B), W.L.O.G. let x ∈ bd(A). If x ∈ A B, then
S
done. Otherwise x ∈ M \(A B). Since x ∈ bd(A), choose ε > 0, there
S
is a y ∈ A ⊆ A B, such that y ∈ D(x.ε) and x ∈ D(x.ε). Hence x ∈
S
S
S S
bd(A B). So x ∈ bd(A B) A B
(c) bd(A
S
B) ⊂ bd(A)
S
(d) bd(bd(bd(A))) = bd(bd(A)).
Solution: bd(bd(bd(A))) is a subset of bd(bd(A)) by (b).
Conversely, for all x ∈ bd(bd(A)), choose ε > 0, there is a y ∈ M \
bd(A) ⊆ M \ bd(bd(A)). Then x, y ∈ D(x, ε), so x ∈ bd(bd(bd(A))).
16. Let a1 =
√
√
√
2, a2 = ( 2)a1 , . . . , an+1 = ( 2)an . Show that an → 2 as n → ∞.
6
(You may use any relevant facts from calculus.)
Solution: It is not difficulty, and I omit it.
17. If
P
xm converges absolutely in Rn , show that
P
xm sin m converges.
P
Solution: Since |xm sin m| ≤ |xm | for all i, and |xm | converges. By Ratio
P
comparison test, |xm sin m| converges.
18. If x, y ∈ M and x 6= y, then prove that there exist open set U and V such that
T
x ∈ U , y ∈ V , and U V = ∅.
Solution: Choose U = D(x, d(x,y)
), V = D(x, d(x,y)
). It is easy to see that U, V
3
3
T
are both open and x ∈ U , y ∈ V . If z ∈ U V , then d(x, y) ≤ d(x, z)+d(y, z) <
T
2d(x,y)
,
then
we
get
a
contradiction.
So
U
V = ∅.
3
19. Define a limit point of a set A in a metric space M to be a point x ∈ M such
T
that U A 6= ∅ for every neighborhood U of x.
(a) What is the difference between limits points and accumulation points?
Give examples.
T
Solution: x is a limit point then (U A)\{x} may be empty set. x is an
T
accumulation point then (U A)\{x} =
6 ∅. Let A = {0}, then limit point
of A is 0, but A has no accumulation points.
(b) If x is a limit point of A, then show that there is a sequence xn ∈ A with
xn → x.
Solution: Given ε1 = 1 > 0 , there is a point x1 ∈ A, such that 0 <
d(x, x1 ) < 1. Let ε2 = d(x, x1 ), there is a point x2 such that 0 < d(x, x2 ) <
d(x, x1 ) = ε2 . Let ε3 = d(x, x2 ), repeat the progress, we will have sequence
xn and xi 6= xj if i 6= j and xn → x.
(c) If x is an accumulation point of A, then show that x is a limit point of A.
Is converse true?
7
Solution: By definition. No, consider A = {0}.
(d) If x is a limit point of A and x ∈
/ A, then show that x is an accumulation
point.
Solution: By definition.
(e) Prove: A set is closed iff it contains all of its limit points.
Solution: By (c) and (d), it is ok.
20. For a set A in a metric space M and x ∈ M , let
d(x, A) = inf{d(x, y) | y ∈ A},
and for ε > 0, let D(A, ε) = {x | d(x, A) < ε}.
(a) Show that D(A, ε) is open.
Solution: If x ∈ D(A, ε)\A, there must be an y 6= x ∈ A, such that
d(x, y) < ε. Pick ε − d(x, y) > 0, for all z ∈ D(x, ε − d(x, y)), d(z, A) ≤
d(z, y) ≤ d(x, z) + d(x, y) < ε − d(x, y) + d(x, y) = ε.
T
If x ∈ D(A, ε) A, for all z ∈ D(x, ε), then d(z, A) ≤ d(z, x) < ε.
(b) Let A ⊂ M and Nε = {x ∈ M | d(x, A) ≤ ε}, where ε > 0. Show that Nε
T
is closed and that A is closed iff A = {Nε | ε > 0}.
Solution: For all accumulation point x of Nε , claim x ∈ Nε . Since x is
an accumulation point of Nε , for each δ > 0, there is a y ∈ Nε , such that
y ∈ D(x, δ). Then d(x, A) ≤ d(x, z) ≤ d(x, y) + d(y, z) < δ + ε. Since δ is
arbitrary, d(x, A) ≤ ε. So we complete the first statement.
T
Since Nε is closed for all ε, then A = {Nε | ε > 0} is closed.
T
Conversely A is closed and A ⊆ Nε for all ε > 0, then we have A ⊆ {Nε |
T
ε > 0}. For x ∈ {Nε | ε > 0}, 0 ≤ d(x, A) ≤ ε. Since ε is arbitrary, then
8
inf d(x, y) = 0 for all y ∈ A. Then x is an accumulation point of A. But
A is closed, so x ∈ A.
21. prove that a sequence xk in a normed space is a Cauchy sequence iff for every
neighborhood U of 0, there is an N such that k, ` ≥ N implies xk = x` ∈ U .
Solution: For every open set U contains 0, there exists ε > 0, such that D(0, ε) ⊆
U . Since xk is a Cauchy sequence, there is a positive integer N , for k, ` ≥ N ,
such that k xk − xl k< ε. Then xk − xl ∈ D(0, ε) ⊆ U .
Conversely given open set D(0, ε), where ε > 0. There is a positive integer N ,
such that k, ` ≥ N implies xk − xl ∈ D(0, ε). So k xk − xl k< ε, then xk is a
Cauchy sequence.
22. Prove Proposition 2.3.2. (Hint: Use Exercise 12 of the Introduction.)
Solution: We can use Proposition 2.1.3 or induction. I think it is not difficulty
to prove it, then I omit it.
23. Prove that the interior of a set A ⊂ M is the union of all the subsets of A that
are open. Deduce that A is open iff A = int(A). Also, give a direct proof of the
latter statement using the definitions.
Solution: We need to proof is that int(A) =
S
{Ai | Ai is an open subset of
A}. Since int(A) is an open subset of A, then LHS⊆ RHS. Given x ∈ RHS,
there exists D(x, ε) ⊆ Ai ⊆ A, hence x ∈ int(A). A is open and A ⊆ A, so
A ⊆ int(A), but int(A) ⊆ A is always true, then A = int(A). If A =int(A) ,
and int(A) is open, so A is open. Suppose A is open, for x ∈ A, there exists an
ε > 0 such that D(x, ε) ⊆ A, since D(x, ε) is an open set, then x ∈ int(A), and
int(A) ⊆ A, so int(A) = A. Suppose A =int(A), and use Exercise 5, then we
complete proof.
9
24. Identify Rn+m with Rn ×Rm . Show that A ⊂ Rn+m is open iff for each (x, y) ∈ A,
with x ∈ Rn , y ∈ Rm , there exist open set U ∈ Rn , V ∈ Rm with x ∈ U , y ∈ V
such that U × V ⊂ A. Deduce that the product of open sets is open.
Solution: Suppose A ⊆ Rn+m is open, for (x, y) ∈ A, there exists an ε > 0, such
that D((x, y), ε) ⊆ A. Let D(x, 2ε ) be U and D(y, 2ε ) be V . Claim U × V ⊆
1/2
D((x, y), ε). For x0 ∈ U , y 0 ∈ V , d((x, y), (x0 , y 0 )) = (d(x, x0 )2 + d(y, y 0 )2 )
1/2
(( 2ε )2 + ( 2ε )2 )
=
√ε
2
<
< ε. So U × V ⊆ A.
On the other hand, given (x, y) ∈ A, there exists D(x, ε) ⊆ U , and D(y, ε) ⊆
V , for some ε > 0. Claim D((x, y), ε) ⊆ U × V . For (x0 , y 0 ) ∈ D((x, y), ε),
d(x, x0 ) ≤ d((x, y), (x0 , y 0 )) < ε. So x0 ∈ D(x, ε) ⊆ U , similarly y 0 ∈ V . Hence
(x0 , y 0 ) ∈ U × V , thus A is open.
Claim U ×V is an open set if U and V are both open sets. Given (x, y) ∈ U ×V ,
there exists D(x, ε) ⊆ U and D(y, ε) ⊆ V for some ε > 0. And we know
D((x, y), ε) ⊆ U × V . So U × V is open.
25. Prove that a set A ⊂ M is open iff we can write A as the union of some family
of ε-disks.
Solution: Suppose A is open, for each x ∈ A, there exists an εx > 0 such
S
S
that A =
D(x, εx ). Since x ∈ A, then x ∈ D(x, εx ) ⊆
D(x, εx ). And
x∈A
x∈A
S
D(x, εx ) ⊆ A for all x, then
D(x, εx ) ⊆ A. Suppose A is the union of some
x∈A
family of ε-disks. And we know ε-disk is open and union of open sets is still
open, thus A is open.
26. Define the sequence of numbers an by
a0 = 1, a1 = 1 +
1
1
, · · · , an = 1 +
.
1 + a0
1 + an−1
Show that an is a convergent sequence. Find the limit.
Solution: It is easy to check that 0 ≤ an ≤ 2 for all n. Now we consider the
subsequence a2n and a2n+1 . We will show the former is strictly increasing and
10
the latter is strictly decreasing. Since
an+1 − an−1 =
−(an − an−2 )
an−1 − an−3
=
(1 + an )(1 + an−2 )
something > 1
This shows an+1 −an−1 has the same sign as an−1 −an−3 . Inductively, an+1 −an−1
has the same sign as a2 − a0 or a3 − a1 . In fact, the second equality above shows
an+1 −an−1 and an −an−2 have opposite signs. Computing the first few terms, it
follows that a2n is strictly increasing while a2n+1 is strictly decreasing. Thus each
of these subsequences converges, and they both obey the recurrence relation
an+1 = 1 +
1
2+
1
1+an−1
Now that we have established that the subsequences converge, it makes sense
to take limits of the above equation.
Since both subsequences satisfy the
same recurrence relation,they converge to the same value(so a converges to
√
this value).We easily compute that lim an = 2.
27. Suppose an ≥ 0 and an → 0 as n → ∞. Given any ε > 0, show that there is a
∞
P
subsequence bn of an such that
bn < ε.
n=1
Solution: Choose ani <
ε
,
2i
it is possible since an → 0. Let ani = bi , then we
complete the proof.
28. Given examples of:
(a) An infinite set in R with no accumulation points
Solution: Z.
(b) A nonempty subset of R that is a contained in its set of accumulation
points
Solution: Q.
(c) A suset of R that has infinitely many accumulation points but contains
none of them
Solution: {2n +
1
m
| n ∈ Z, m = 2, 3, . . . }
11
(d) A set A such that bd(A) =cl(A).
Solution: Q.
You may check it by yourself.
29. Let A, B ⊂ Rn and x be an accumulation point of A
S
B. Must x be an
accumulation point of either A or B?
Solution: Yes. If x is not an accumulation point of neither A nor B, then
T
there is some ε > 0, such that (D(x, ε)\{x}) A = ∅ and ε0 > 0, such that
T
(D(x, ε0 )\{x}) B = ∅. Then choose δ = min{ε, ε0 } such that
(D(x, δ)\{x})
\
A ⊂ (D(x, ε)\{x})
\
A=∅
(D(x, δ)\{x})
\
B ⊂ (D(x, ε)\{x})
\
B=∅
and
T S
S
Hence (D(x, δ)\{x}) (A B) = ∅, so x is not an accumulation point of A B,
a contradiction.
30. Show that any open set in R is a union of disjoint open intervals. Is this sort of
result true in Rn for n > 1, where we define an open interval as the Cartesian
product of n open intervals, ]a1 , b1 [× · ×]an , bn [?
Solution: You may read Theorem 3.11 at page 51 of Mathematical Analysis, 2nd, Apostol.
No. Consider the set (0, 2) × (0, 2) \ [1, 2) × [1, 2) .
31. Let A0 denote the set of accumulation points of a set A. Prove that A0 is closed.
Is (A0 )0 = A0 for all A?
Solution: Suppose x is an accumulation points of A0 , then given ε > 0, there is
a y ∈ A0 such that d(x, y) < 2ε . Since y is an accumulation points of A, there is
a z ∈ A, such that d(z, y) < 2ε . So there is a z ∈ A, such that
d(z, x) ≤ d(x, y) + d(z, y) <
12
ε ε
+ = ε.
2 2
Then x ∈ A0 , so A0 is closed.
No. Take A = {1/n | n = 1, 2, . . . }, but A0 = {0}, (A0 )0 = ∅.
32. Let A ⊂ Rn be closed and xn ∈ A be a Cauchy sequence. Prove that xn
converges to a point in A
Solution: Since A is closed, choose a subsequence of xn converges to a point
x ∈ A. Since a subsequence of a Cauchy sequence converges to a point, then
the Cauchy sequence converges to the point. Hence xn converges to the point
x ∈ A.
33. Let sn be a bounded sequence of real numbers. Assume that 2sn ≤ sn−1 + sn+1 .
Show that lim (sn+1 − sn ) = 0.
n→∞
Solution: It is easy to see that sn − sn−1 is an increasing sequence. If there
exists si+1 − si = a > 0, then sn+i ≥ si + na. So lim sn = ∞. So sn+1 − sn ≤ 0
n→∞
for all n, then sn is a decreasing sequence. Since R is a complete ordered field,
then lim (sn+1 − sn ) = 0.
n→∞
34. Let xn ∈ Rk and d(xn+1 , xn ) ≤ rd(xn , xn−1 ), where 0 ≤ r < 1. Show that xn
converges.
Solution: Tt is easy to check that xn is a Cauchy sequence, thus it converges.
35. Show that any family of disjoint nonempty open sets of real numbers is countable.
Solution: Since every open set Ai of real numbers contains a rational number,
S
ri . And Q is countable, then we have that ri is countable, so we complete
the proof.
36. Let A, B ⊂ Rn be closed sets. Does A + B = {x + y | x ∈ A, y ∈ B} have to be
closed?
13
Solution: No. Take A = {−1, −2, −3, . . . } and B = {1 + 12 , 2 + 13 , 3 + 41 , . . . }
Thus 0 is accumulation point of A + B, but 0 ∈
/ A + B.
37. For A ⊂ M , a metric space, prove that
bd(A) = [A
\
cl(M \A)]
[
[cl(A)\A]
T
cl(M \A), then if x ∈ bd(A), x ∈ cl(A) and
T
x ∈ cl(M \A). If x ∈ A, then x ∈ A cl(M \A). If x ∈
/ A, then x ∈ cl(A)\A.
Solution: Since bd(A) = cl(A)
Hence LHS ⊆ RHS.
T
T
If x ∈ A cl(M \A), then x ∈ cl(A) cl(M \A). If x ∈ cl(A)\A, then x ∈ M \A.
T
So x ∈ cl(M \A), hence x ∈ cl(A) cl(M \A). Thus RHS ⊆ LHS.
38. Let xk ∈ Rn satisfy kxk − x` k ≤ 1/k + 1/`. Prove that xk converges.
Solution: For all ε > 0 , there is an integer N > 2ε , such that for k, ` > N ,
kxk − x` k ≤ 1/k + 1/` < 2/N < ε. Then xk converges.
39. Let S ⊂ R be bounded above and below. Prove that sup(S) − inf(S) = sup{x −
y | x ∈ S and y ∈ S}.
Solution: For all x, y ∈ S, x ≤ sup(S), −y ≤ − inf(S), then x − y ≤ sup(S) −
inf(S). So
sup{x − y | x ∈ S and y ∈ S} ≤ sup(S) − inf(S).
For all ε > 0, there exists x and y such that sup(S) − ε/2 < x, − inf(S) − ε/2 <
−y. So
sup(S) − inf(S) − ε < x − y ≤ sup{x − y | x ∈ S and y ∈ S}.
Since ε is arbitrary, then
sup(S) − inf(S) ≤ sup{x − y | x ∈ S and y ∈ S}
.
14
40. Suppose in R that for all n, an ≤ bn , an ≤ an+1 , and bn+1 ≤ bn . Prove that an
converges.
Solution: Claim an is bounded. Since a0 ≤ an ≤ bn ≤ b0 for all n, then an is
bounded and R is a complete ordered field. So an converges.
41. Let An be subsets of a metric space M , An+1 ⊂ An , and An 6= ∅, but assume
∞
∞
T
T
that
An = ∅. Suppose x ∈
cl(An ). Show that x is an accumulation point
n=1
n=1
of A1 .
Solution: Suppose x ∈ A1 but not accumulation point of A1 . Then x must be an
accumulation point of Ak and x ∈
/ Ak , for some k > 1. However it is impossible
because Ak ⊂ A1 . Then x must be an accumulation point of A, since x ∈ cl(A).
42. Let A ⊂ Rn and x ∈ Rn . Define d(x, A) = inf{d(x, y) | y ∈ A}. Must there be
a z ∈ A such that d(x, A) = d(x, z)?
Solution: No. Let A = {1/n | for n ∈ N}. Choose x = 0, d(x, A) = 0, but
there is no z ∈ A such that d(x, z) = 0.
43. Let x1 =
√
3, . . . , xn =
√
3 + xn−1 . Compute lim xn .
n→∞
Solution: It is easy then I omit it.
44. A set A ⊂ Rn is said to be dense in B ⊂ Rn if B ⊂ cl(A). If A is dense in Rn
T
and U is open, prove that A U is dense in U . Is this true if U is not open?
S
Solution: Since A is dense in Rn , Rn ⊂ cl(A) = A acc(A), for all x ∈ U ⊂ R,
T
T
if x ∈ A, then x ∈ A U ⊂ cl(A U ), then done.
Otherwise, x ∈ acc(A). We know for all ε > 0, there is an y 6= x, y ∈ A such
that y ∈ D(x, ε). Since A is open there is an ε0 > 0, D(x, ε0 ) ⊂ U then for
T
all εi ≤ ε0 , there is a point yi ∈ A U and let ε0 = ε. Otherwise, choose
T
y0 ∈ D(x, ε) ⊂ D(x, εi ). So x ∈ acc(A U )
15
No, Q is dense in Rn , choose U = {π}. Then U = {π}, and A
T
cl(A U ).
T
U = ∅ =
45. Show that x( log x) = o(ex ) as x → ∞ (see Worked Example 2.5).
Solution: It’s easy, then I omit it.
46. (a) If f = o(g) and if g(x) → ∞ as x → ∞, then show that ef (x) = o(eg(x) ).
Solution: Since (g − f )/g → 1 as x → ∞, then e(g−f )(x) ∼ eg(x) = ∞ as
x → ∞, so e(f −g)(x) = 0 as x → ∞. Hence ef (x) = o(eg(x) ).
(b) Show that lim (x log x)/ex = 0 by showing that x = o(ex/2 ) and that
x→∞
x/2
log x = o(e
).
Solution: It is easy to proof that x = o(ex/2 ) and log x = o(ex/2 ). Using
lim f (x)g(x) = lim f (x)
lim g(x) , whenever lim f (x) and lim g(x)
x→∞
x→∞
x→∞
are both finite as x → ∞.
47. Show that γ = lim
n→∞
1 + 1/2 + · · · + 1/n − log n exists by using the proof of
the integral test (γ is called Euler’s constant).
Solution: Let an = 1 + 1/2 + · · · + 1/n − log n , then an+1 − an = 1/(1 + n) −
log(n + 1) + log n < 0 and it is clearly that an > 0 (by graph), so an is converge.
Thus γ exists.
48. (a) If an > 0 and lim sup an+1 /an < 1, then
n→∞
P
1, then
an diverges.
(b) If an ≥ 0 and if lim sup
√
n
P
an converges, and if lim inf an+1 /an >
an < 1 (respectively, ¿1), then
n→∞
P
an converges
n→∞
(respectively, diverges).
(c) In the ratio comparison test, can the limits be replaced by lim sup’s.
Solution: a. and b. You may read p.65
p.67, Principle of Mathmatical
Analysis, 3nd, Rudin. However, I don’t know how to solve c.
16
49. Prove Raabe’s test: If an > 0 and if an+1 /an ≤ 1 − A/n for some fixed
P
constant A > 1 and n sufficiently large, then
an converges. Similarly, show
P
that if an+1 /an > 1 − (1/n), then
an diverges.
Use Raabe’s test to prove convergence of the textbfhypergeometric series whose
general term is
an =
α(α + 1) · · · (α + n − 1)β(β + 1) · · · (β + n − 1)
1 · 2 · · · n · γ · (γ + 1) · · · (γ + n − 1)
where α, β, and γ are nonnegative integers, γ > α + β. Show that the series
diverges if γ < α + β.
Solution: I don’t know how to prove Raabe’s test.
Use Raabe’s test. Note that an+1 /an = (α + n)(β + n)/(n + 1)(n + γ), then it
is easy to check.
2
50. Show that for x sufficiently large, f (x) = (xcos2 x + sin2 x)ex is monotonic and
2
tends to +∞, but that neither the ratio f (x)/(x1/2 ex ) nor its reciprocal is
bounded.
√
Solution: It is easy to check f (x) monotone and tends to ∞. Since 1/ x ≤
√
√
2
f (x)/x1/2 ex ≤ x and 1/ x is not bounded. Then we complete the proof. 51. (a) If un > 0, n = 1, 2, . . . show that
lim inf
√
√
un+1
un+1
≤ lim inf n un ≤ lim sup n un ≤ lim sup
.
un
un
Solution: You may read p.68 ∼ p.69, Principle of Mathmatical
Analysis, 3nd, Rudin.
(b) Deduce that if lim(un+1 /un ) = A, then lim sup
√
n
un = A.
Solution: It is trivial, since lim sup = limit = lim inf.
(c) Show that the converse of part b is false by use of the sequence u2n =
u2n+1 = 2−n .
Solution: You can check it by yourself.
17
(d) Calculate lim sup
√
n
n!/n.
Solution: Let un = n!/nn , and easy calculate lim un+1 /un .
n→∞
52. Test the following series for convergence.
(a)
(b)
(c)
(d)
(e)
(f)
∞
P
n=0
∞
P
n=0
∞
P
n=0
∞
P
n=0
∞
P
n=0
∞
P
n=0
−k
√e
k+1
k
k2 +1
√
Solution: converge
Solution: diverge
n+1
n2 −3n+1
Solution: converge
log(k+1)−log k
tan−1 (2/k)
Solution: diverge
sin(nα ), α real, ¿ 0 Solution: converge for α > 1; diverge for 0 < α ≤ 1
n3
3n
Solution: converge
It is easy to check, then I omit it.
53. Given a set A in a metric space, what is the maximum number of distinct subsets
that can be produced by successively applying the operation closure, interior,
and complement to A (in any order)? Give an example of a set achieving your
maximum.
Solution: The answer is 14. A, int(A), cl(A), cl(int(A)), int(cl(A)), int(cl(int(A))),
cl(int(cl(A))) Ac , int(Ac ), cl(Ac ), cl(int(Ac )), int(cl(Ac )), int(cl(int(Ac ))), cl(int(cl(Ac ))).
Just only need to check the following two statements.
cl(int(cl(int(A)))) = cl(int(A)) . . . . . . (1)
proof of (1): It is easy to see LHS ⊂ RHS. Conversely claim int(A) ⊂
int(cl(int(A))). Suppose x ∈ int(A), there is a D(x, ε) ⊂ A, for some ε > 0.
Then D(x, ε) ⊂ int(A), since D(x, ε) is an open subset of A. So D(x, ε) ⊂
18
cl(int(A)), hence x ∈ int(cl(int(A))). Then we have cl(int(A)) ⊂ cl(int(cl(int(A)))).
Then we complete the proof.
int(cl(int(cl(A)))) = int(cl(A)) . . . . . . (2)
proof of (2): It is easy to see that RHS ⊂ LHS. Conversely claim cl(int(cl(A))) ⊂
cl(A). Suppose x ∈ int(cl(A)), then x ∈ cl(A). Otherwise x is an accumulation
point of int(cl(A)). Thus for all ε > 0, there exist y ∈ int(cl(A)) ⊂ cl(A), such
that y ∈ D(x, ε), then x is an accumulation point of cl(A). But cl(A) is closed,
then we have that x ∈ cl(A). So we have int(cl(int(cl(A)))) ⊂ int(cl(A)).
Let the metric space M = R. Then the following sets are desired.
S
S
S T
A = [0, 1) (1, 2] {3} {Q [4, 5]}
S
int(A) = (0, 1) (1, 2)
S
S
cl(A) = [0, 2] {3} [4, 5]
cl(int(A)) = [0, 2]
S
int(cl(A)) = (0, 2) (4, 5)
int(cl(int(A))) = (0, 2)
S
cl(int(cl(A))) = [0, 2] [4, 5]
S
S
S
S
T
S
Ac = (−∞, 0) {1} (2, 3) (3, 4) {Q0 [4, 5]} (5, ∞)
S
S
S
int(Ac ) = (−∞, 0) (2, 3) (3, 4) (5, ∞)
S
S
cl(Ac ) = (−∞, 0] {1} [2, ∞)
S
S
cl(int(Ac )) = (−∞, 0] [2, 4] [5, ∞)
S
int(cl(Ac )) = (−∞, 0) (2, ∞)
S
S
int(cl(int(Ac ))) = (−∞, 0) (2, 4) (5, ∞)
S
cl(int(cl(Ac ))) = (−∞, 0] [2, ∞)
19