MATH 6337: Homework 1
1.1. Prove the following facts, which were left as exercises.
(a) For a sequence of sets {Ek }, lim sup Ek consists of those points which belong to infinitely
many Ek , and lim inf Ek consists of those points which belong to all Ek from some k on.
S
Solution. Suppose x ∈ lim sup Ek . Then x ∈ k≥j Ek for all j, so for every j ≥ 1, there
exists k ≥ j such that x ∈ Ei ; hence x ∈ Ek for infinitely many k. All of these implications
are reversible.
T
Suppose x ∈ lim inf Ek . Then x ∈ k≥j Ek for some k. Thus, for some j ≥ 1, x ∈ Ek for all
k ≥ j; hence x ∈ Ek for all but finitely many k. All of these implications are reversible. (b) The De Morgan laws:
!c
[
E
!c
=
E∈F
S
\
Ec
and
E∈F
c
\
E∈F
E
=
[
E c.
E∈F
S
/ E∈F E, so x fails to be in every E ∈ F; hence
Solution. Suppose x ∈
E∈F E ; then x ∈
T
c
c
x ∈ E for every E ∈ F, so x ∈ E∈F E . All of these implications are reversible.
c
T
T
/ E∈F E, so x fails to be some E ∈ F; hence x ∈ E c
Suppose x ∈
E∈F E ; then x ∈
S
for some E ∈ F, so x ∈ E∈F E c . All of these implications are reversible.
[Note: F is an arbitrary family of sets and may have infinite size, so a proof by induction
cannot work here.]
(d) Theorem 1.4:
(a) L = lim supk→∞ ak if and only if (i) there is a subsequence akj of {ak } which
converges to L and (ii) if L0 > L, there is an integer K such that ak < L0 for k ≥ K.
(b) ` = lim inf k→∞ ak if and only if (i) there is a subsequence akj of {ak } which
converges to ` and (ii) if `0 < `, there is an integer K such that ak > `0 for k ≥ K.
Solution.
(a) First, the forward direction. Let Lj = supk≥n ak , so L = limn→∞ Ln . Given j ≥ 1,
choose nj ≥ j such that Lnj − L < 2j1 and choose kj ≥ nj such that akj − Lnj < 2j1 .
Then
ak − L ≤ ak − Ln + Ln − L < 1 ,
j
j
j
j
j
0
so akj → L. Now, given L > L, choose K such that |LK − L| < L0 − L; then
supk≥K ak < L0 , so ak < L0 for all k ≥ K.
Now, the reverse direction. Let ε > 0; then there exists J such that for every
j ≥ J, akj − L < ε. Thus, supk≥J ak = LJ ≥ L − ε; thus, limj→∞ Lj ≥ L.
On the other hand, there exists K such that for every k ≥ K, ak < L + ε; thus,
supk≥K ak = LK ≤ L + ε, so limj→∞ Lj ≤ L. Thus, lim sup ak = limj→∞ Lj = L.
(b) Recall that lim inf ak = − lim sup(−ak ). Using the result of part (a), if ` = lim inf ak ,
then −` = lim sup(−ak ), so there is a subsequence −akj of {−ak } converging to
−` (thus a subsequence akj of {ak } converging to `), and if `0 < `, then −`0 > `,
so there exists K such that −ak < −`0 (hence ak > `0 ).
On the other hand, given {ak }, if there is a subsequence akj converging to ` and
for all `0 < ` there exists K such that ak > `0 whenever k ≥ K, then the hypothesis
of the reverse implication of part (a) is satisfied: there is a subsequence −akj of
{−ak } which converges to −` and, if −`0 > −`, there is an integer K such that
−ak < −`0 for k ≥ K. Thus −` = lim sup(−ak ), so ` = lim inf ak .
(n) Theorem 1.14:
(a) M = lim supx→x0 ;x∈E f (x) if and only if (i) there exists {xk } in E such that xk → x0
and f (xk ) → M , and (ii) if M 0 > M , there exists δ > 0 such that f (x) < M 0 for
x ∈ B 0 (x0 ; δ) ∩ E.
(b) m = lim inf x→x0 ;x∈E f (x) if and only if (i) there exists {xk } in E such that xk → x0
and f (xk ) → m, and (ii) if m0 < m, there exists δ > 0 such that f (x) > m0 for
x ∈ B 0 (x0 ; δ) ∩ E.
Solution. For simplicity, let Bδ denote the open ball of radius δ centered at x0 intersected
with E.
(a) First, the forward direction. Let Mδ = supBδ f (x), so M = limδ→0 Mδ . Given
1
k ≥ 1, choose δk ≤ 1j such that |Mδk − M | < 2k
, and choose xk ∈ Bδk such that
1
|f (xk ) − Mδk | < 2k . Then
|f (xk ) − M | ≤ |f (xk ) − Mδk | + |Mδk − M | <
1
,
k
so f (xk ) → M (and xk → x0 since |xk − x0 | ≤ k1 ). Now, given M 0 > M , choose δ
such that |Mδ − M | < M 0 − M ; then supBδ f (x) < M 0 , so f (x) < M 0 for all x ∈ Bδ .
Now, the reverse direction. Let ε > 0; then there exists K such that for every
k ≥ K, |f (xk ) − M | < ε. Thus, letting δk = |xk − x0 |, we have supBδ f (x) = Mδk ≥
k
M − ε; thus, limδ→0 Mδ ≥ M . On the other hand, there exists δ such that for every
x ∈ Bδ , f (x) < M + ε; thus, supBδ f (x) = Mδ ≤ M + ε, so limδ→0 Mδ ≤ M . Thus,
lim sup f (x) = limδ→0 Mδ = M .
(b) Recall that lim inf f (x) = − lim sup(−f (x)). Using the result of part (a), if m =
lim inf f (x), then −m = lim sup(−f (x)), so there is a sequence {−f (xk )} (with
xk → x0 ) converging to −m (thus a sequence {f (xk )} with xk → x0 and f (xk ) → m),
and if m0 < m, then −m0 > m, so there exists δ such that −f (x) < −m0 (hence
f (x) > m0 ) for all x ∈ Bδ .
On the other hand, if there is a sequence {f (xk )} with xk → x0 and f (xk ) → m
and for all m0 < m there exists δ such that f (x) > m0 whenever x ∈ Bδ , then the
hypothesis of the reverse implication of part (a) is satisfied: there is a sequence with
xk → x0 and −f (xk ) → −m and, if −m0 > −m, there exists δ such that −f (x) < −m0
for x ∈ Bδ . Thus −m = lim sup(−f (x)), so m = lim inf f (x).
1.2. Find lim sup Ek and lim inf Ek if Ek = [−1/k, 1] for k odd and Ek = [−1, 1/k] for k
even.
Solution. Recall that lim sup Ek is the set of points which appear in infinitely many Ek , and
lim inf Ek is the set of points which appear in all but finitely many Ek . If x ∈ [−1, 0], then
x ∈ Ek for all even k, and if x ∈ [0, 1], then x ∈ Ek for all odd k, so lim sup Ek = [−1, 1]. If
x ∈ [−1, 0), then x ∈
/ Ek for odd k when x < −1/k, which must eventually happen for large
enough k; similarly, if x ∈ (0, 1], then x ∈
/ Ek for even k when x > 1/k. Thus, every point
but x = 0 fails to appear in infinitely many Ek , so lim inf Ek = {0}.
1.3.
(a) Show that (lim sup Ek )c = lim inf Ekc .
(b) Show that if Ek % E or Ek & E, then lim sup Ek = lim inf Ek = E.
Solution.
(a) This follows from the De Morgan laws:
!!c
!c
∞
∞
∞
[
[
\
[
[
c
Ek
=
Ek =
(lim sup Ek ) =
j=1
k≥j
j=1
k≥j
!
\
j=1
Ekc
= lim inf Ekc .
k≥j
S
(b) Ek % E means that Ek ⊆ Ek+1 and Ek = E; in this case,
!
!
∞
∞
∞
∞
[
\
[
\
[
\
lim inf Ek =
Ek =
Ej = E and lim sup Ek =
Ek =
E = E.
j=1
k≥j
j=1
j=1
k≥j
j=1
T
On the other hand, Ek & E means that Ek ⊇ Ek+1 and Ek = E; in this case,
!
!
∞
∞
∞
∞
[
\
[
\
[
\
lim inf Ek =
Ek =
E = E and lim sup Ek =
Ek =
Ej = E.
j=1
k≥j
j=1
j=1
k≥j
j=1
1.9. Prove that any closed subset of a compact set is compact.
Solution. Let X be a compact space, let F ⊆ X be closed in X, and let C = {Gα } be an
open cover of F ; then each Gα = G0α ∩ F for a subset G0α ⊆ X which is open in X. Since
S
S
X = F ∩ F c and G0α ⊇ F , F c ∪ G0α = X; moreover, since F is closed in X, F c is open
in X. Thus, C 0 = {G0α } ∪ {F c } is an open cover of X.
Since X is compact, there exists a finite subcover C 00 = {G01 , ..., G0n } ⊆ C 0 such that
Sn
Sn
0
0
0
0
k=1 Gk ⊇ F . Now, each Gk has the property that either Gk ∩ F =
k=1 Gk = X; hence
Gk ∈ C or Gk ∩ F = ∅ (if Gk = F c ). Thus, we can take C 000 = {G1 , G2 , ..., Gn } \ {∅}; then
S
000
⊂ C, so C 000 is a finite open subcover of C. Hence F is a compact
G∈C 000 G = F and C
subspace of X.
[Note that X need not have a norm, so arguments involving boundedness would fail in
such a case.]
1.16. If {fk } is a sequence of bounded, Riemann-integrable functions on an interval I which
converges uniformly on I to f , show that f is Riemann integrable on I and that
ˆ
ˆ
(R) fk (x) dx → (R) f (x) dx.
I
I
Solution. Given ε > 0, choose k such that for every x ∈ I, we have
|fk (x) − f (x)| < ε;
it follows that
sup fk (x) − sup f (x) < ε and
I
I
inf fk (x) − inf f (x) < ε.
I
I
Since fk is Riemann-integrable, there exists a partition Γ such that
UΓ (fk ) − LΓ (fk ) < ε.
Then
|UΓ (f ) − LΓ (f )| ≤ |UΓ (f ) − UΓ (fk )| + |UΓ (fk ) − LΓ (fk )| + |LΓ (fk ) − LΓ (f )| ≤ 3ε |I| .
Hence f is also Riemann-integrable. Moreover,
ˆ
ˆ
ˆ
(R) fk (x) dx − (R) f (x) dx ≤ (R) |fk (x) − f (x)| dx < ε |I| .
I
I
I
Hence the integral of f is the limit of the integrals of the fk .