The University of Sydney
School of Mathematics and Statistics
Solutions to Tutorial 7 (Week 8)
MATH2962: Real and Complex Analysis (Advanced)
Semester 1, 2017
Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/
Lecturer: Florica C. Cı̂rstea
Questions marked with * are more difficult questions.
Material covered
(1) Topology of RN and CN .
(2) Limits and properties of continuous functions
Outcomes
This tutorial helps you to
(1) be familiar with the basic topological properties of the real and complex euclidean space;
(2) have solid foundations in the more formal aspects of analysis, including a knowledge of
precise definitions, how to apply them and the ability to write simple proofs;
Summary of essential material
The set
B(x, r) := {y ∈ KN : ky − xk < r}.
is called the open ball centred at x with radius r > 0.
• U ⊆ RN is called open if either U = ∅ or for every x ∈ U if there exists r > 0 such that
B(x, r) ⊆ U
• A ⊆ KN is called closed if the complement Ac = {x ∈ RN : x 6∈ A} is open.
In the context of continuous function we are also interested in sets that are open or closed with
respect to some subset D ⊆ Rd :
• U ⊆ D is called open in D if either U = ∅ or for every x ∈ U if there exists r > 0 such
that B(x, r) ∩ D ⊆ U
• A ⊆ KN is called closed if the complement Ac ∩ D = {x ∈ RN : x 6∈ A} is open in D.
Open or relatively open sets have the following properties:
• The whole set and the empty set are open;
• An arbitrary union of open sets is an open set;
• A finite intersection of open sets are open set.
c 2017 The University of Sydney
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Every set A ⊆ RN is associated with an open and a closed set:
• The interior of A is open:
int(A) := {x ∈ A : there exists r > 0 so that B(x, r) ⊆ A}.
int(A) ⊆ A with equality if and only if A is open.
• The closure of A is closed:
Ā := {x ∈ RN : B(x, r) ∩ A 6= ∅ for all r > 0};
A ⊆ Ā with equality if and only if A is closed.
• The boundary of A: ∂A = Ā \ int(A).
Sequential characterisations:
• x ∈ A if and only if there exists a sequence (xn ) in A such that xn → x
as n → ∞.
• A is bounded if and only if every sequence in A has a convergent
subsequence.
• A is closed and bounded if and only if every sequence in A has a
convergent subsequence with limit in A.
Such a set is also called (sequentially) compact.
Continuous functions can be characterised by means of open or closed sets. Let D ⊆ Rd and
f : D → RN a function. Given U ⊆ RN (a subset of the codomain of f ), we define the inverse
image or pre-image of U to be the set
f −1 [U ] := {x ∈ D : f (x) ∈ U }.
The following statements are equivalent:
(i) f : D → RN is continuous;
(ii) f −1 [U ] is open in D for every open set U ⊆ RN ;
(iii) f −1 [A] is closed in D for every closed set A ⊆ RN ;
Note however, that the image of an open or closed set does not need to be open or closed,
respectively!
However, if A is closed and bounded (compact) and f is continuous, then also f (A) is closed
and bounded (compact). As a consequence a continuous function on a closed and bounded set
into R has a maximum and a minimum.
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Questions to complete during the tutorial
1. Let f : D → KN be a function. Give a formal definition of the following limits. Include
the relevant assumptions on the domain and co-domain of the function.
(a) lim f (x) = ∞;
x→a
Solution: Suppose that f : D → R and a ∈ D with D ⊂ Kd . Then f (x) → ∞
as x → a if for every M ∈ R, there exists δ > 0 such that f (x) > M for all x ∈ D
with 0 < kx − ak < δ.
(b)
lim f (x) = ∞;
x→−∞
Solution: Suppose that f : D → R with D ⊂ R and D ∩ (−∞, −n) 6= ∅ for all
n ∈ N. Then f (x) → ∞ as x → −∞ if for every M ∈ R there exists x0 ∈ R such
that f (x) > M for all x ∈ D with x < x0 .
(c)
lim f (x) = b.
x→a+
Solution: Suppose that f : D → KN with D ⊂ R and (a, a + r) ∩ D 6= ∅ for
all r > 0. We write lim f (x) = b if for every ε > 0 there exists δ > 0 such that
x→a+
kf (x) − bk < ε for all x ∈ D ∩ (a, a + δ).
2. Decide whether the following sets are open or closed. Determine interior, closure and
boundary.
(a) [1, ∞) ⊆ R;
Solution: The set [1, ∞) is closed since its complement (−∞, 1) is open. We
have int[1, ∞) = (1, ∞) and the closure of [1, ∞) is the set itself. The boundary of
[1, ∞) is the set {1}, which is obtained by subtracting int[1, ∞) from the closure
of [1, ∞).
(b) {(−1)n + 1/n : n ∈ N \ {0}} ⊆ R;
Solution:
The set is neither closed nor open. The closure is the sequence
together with all its points of accumulation. The set consists of the points xn =
(−1)n + 1/n for n ≥ 1. The sequence (xn ), namely the limits 1 and −1 of the
subsequences (x2k ) and (x2k+1 ), respectively. Hence, the closure of the given set is
{(−1)n + 1/n : n ∈ N \ {0}} ∪ {1, −1}. The interior is empty and the boundary is
the same as its closure.
(c) {z ∈ C : |z − 1| + |z + 1| < 4} ⊆ C;
Solution: The function f : C → R given by f (z) := |z − 1| + |z +1| is continuous.
The given set is the inverse image of the open set (−∞, 4), and therefore an open
set. It is the interior of an ellipse with foci at x = ±1 without the boundary. The
closure is the ellipse including the line bounding it, and the boundary is the ellipse
|z − 1| + |z + 1| = 4.
(d) Z ⊆ R;
S
Solution: The complement of Z in R is R \ Z = k∈Z (k, k + 1), which is an open
set (as the union of open sets). This shows that Z is closed. Moreover, we have
int(Z) = ∅ and ∂Z = Z.
(e) {(x, y, z) : x2 + y 2 < 1, x + y + z = 1};
Solution: Geometrically, the set is the part of the plane x + y + z = 1 within
the open cylinder x2 + y 2 < 1. Hence the set is neither open nor closed. Since it
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is part of a plane, its interior is empty. Its closure and boundary is {(x, y, z) ∈
R3 : x2 + y 2 ≤ 1, x + y + z = 1}
(f) {n/(n + 1) : n ∈ N} ∪ {1} ⊆ R.
Solution:
We know that n/(n + 1) → 1 as n → ∞, so the set is closed.
The reason is that every convergent sequence in the set has its limit in the set.
The interior is empty since every interval about a point in the set intersects the
complement of the set. The boundary equals the set itself.
3.
(a) Suppose Kn ⊆ RN are closed non-empty sets such that Kn+1 ⊆ Kn for all n ∈ N
and that
diam(Kn ) := sup kx − yk → 0.
x,y∈Kn
(i) For every n ∈ N let xn ∈ Kn . Show that (xn ) is a Cauchy sequence in RN .
Solution: By assumption, Kn+1 ⊆ Kn for all n ∈ N. Hence, that xm ∈
Km ⊆ Kn for all m > n. As xm , xn ∈ Kn we have
kxm − xn k ≤ sup kx − yk = diam(Kn )
x,y∈Kn
for all m > n. Since diam(Kn ) → 0 as n → ∞, given ε > 0 there exists
n0 ∈ N such that diam(Kn0 ) < ε for all n > n0 . Hence
kxm − xn k ≤ diam(Kn ) < ε
for all m > n > n0 , showing that (xn ) is a Cauchy sequence.
T
(ii) Hence, show that n∈N Kn 6= ∅. (This is called Cantor’s Intersection Theorem.)
Solution: Let (xn ) be the sequence from part (??). Hence (xn ) converges,
that is, xn → x. We know that xn ∈ Km for all n > m. As T
Km is closed
x ∈ Km . Since this is true for all m ∈ N we conclude that x ∈ m∈N Km , so
the intersection is non-empty as claimed.
(b) Give an example of non-empty open
T bounded sets On with On+1 ⊆ On and
diam(On ) → 0 as n → ∞ such that n∈N On = ∅.
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Solution: For every n ∈ N, we defineTOn = (0, n+1
) ⊆ R. Then clearly On
is open, On+1 ⊆ On for all n ∈ N and n∈N On = ∅. Moreover, diam(On ) =
1/(n + 1) → 0 as n → ∞.
The example shows that the requirement that Kn being closed in part (??) is
essential.
4. Let D ⊆ Rm be closed and bounded. Suppose that f : D → Rn is an injective continuous
function. Show that the inverse function f −1 : f (D) → D is continuous.
Solution: In lectures it was proved that a function is continuous if and only if the
inverse image of every closed set is closed. If U ⊆ Rm is closed, then
[f −1 ]−1 [U ] = f (U ∩ D)
is closed. Hence f −1 is continuous.
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Extra questions for further practice
5. If A ⊆ R is a closed set bounded from above (below), show that A has a maximum
(minimum).
Solution: Suppose that A is bounded from above. By the least upper bound axiom
m := sup A exists. Now for every ε > 0 there exists a ∈ A such that m − ε < a ≤ m.
This means that B(m, ε) ∩ A 6 ∅ for all ε > 0. Hence m is in the closure of A. As A is
closed we conclude that m ∈ A.
6. Let X ⊆ KN be an arbitrary set. A set A ⊆ X is said to be relatively open in X if for
every x ∈ A there exists r > 0 such that B(x, r) ∩ X ⊆ A. Moreover, A ⊆ X is called
relatively closed in X if its complement Ac ∩ X is relatively open in X.
(a) Which of the following sets are relatively open or closed in X = [0, 2)?
(i) A = [0, 1)
Solution: The set is relatively open in [0, 2), because about every point x
in [0, 1) there is an open interval whose intersection with [0, 2) is entirely in
[0, 1). Note that Ac ∩ X = [1, 2) is not relatively open in X since 1 ∈ Ac ∩ X
and there is no small interval about 1 entirely contained in Ac ∩ X. Hence,
[0, 1) is not relatively closed in X.
(ii) A = [1, 2)
Solution: The set [1, 2) is relatively closed in X = [0, 2) since its complement in X, namely [0, 1), is relatively open in X. However, [1, 2) is not
relatively open in [0, 2) since [0, 1) is not relatively closed in X.
(iii) A = [1/2, 1)
Solution: The set [1/2, 1) is not relatively open in [0, 2) because there
is no small interval about 1/2 entirely contained in the intersection A ∩ X.
The set A = [1/2, 1) is also not relatively closed in [0, 2) because Ac ∩ X =
[0, 1/2) ∪ [1, 2) is not relatively open in X. Indeed, there is no small interval
about 1 entirely contained in the intersection Ac ∩ X.
(b) Prove that A ⊆ X is relatively open in X if and only if there exists an open set O
in KN with A = X ∩ O.
Solution: First suppose that A is open in X. Then for every x ∈ A there exists
rx > 0 such that B(x, rx ) ∩ X ⊆ A. Since arbitrary unions of open sets are open
the set
[
O :=
B(x, rx )
x∈A
is an open set. Since A ⊆ O and B(x, rx )∩X ⊆ A for all x ∈ A we have A = X ∩O.
Hence there exists an open set as required. Suppose now that O is an open set
with X ∩ O = A. If x ∈ A then x ∈ O and since O is open there exists r > 0 such
that B(x, r) ⊆ O. Hence B(x, r) ∩ X ⊆ O ∩ X = A, showing that A is open in X.
(c) Show that ∅ and X are relatively open in X. Prove that arbitrary unions and
finite intersections of relatively open sets in X are relatively open in X.
Solution: We use the characterisation of relatively open sets from part (??).
Clearly X = X ∩ KN , so X is open in X. Similarly ∅ is open in X since ∅ ∩ X = ∅
and ∅ is open in KN . If Oα , α ∈ A is a family of open sets in X then by part
S (??)
N
there exist open sets Õα ⊆ K with Õα ∩ X = Oα for all α ∈ A. Now α∈A Õα
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S
S
is open in KN , and thus ( α∈A Õα ) ∩ X = α∈A Oα is open in X. The proof for
finite intersections is similar.
7. Let A, B ⊆ RN .
(a) If A ⊆ B, show that A ⊆ B and that int(A) ⊆ int(B).
Solution: Suppose that x ∈ A. Then B(x, r)∩A 6= ∅ for all r > 0. Since A ⊆ B,
we also have B(x, r) ∩ B 6= ∅ for all r > 0. Hence x ∈ B, proving that A ⊆ B.
Suppose now that x ∈ int(A). Then, there exists r > 0 such that B(x, r) ⊆ A,
and thus B(x, r) ⊆ B since A ⊆ B. This means that x ∈ int(B), showing that
int(A) ⊆ int(B).
*(b) Show that A ∪ B = A ∪ B
Solution: Clearly A ∪ B ⊆ A ∪ B. As a finite union of closed sets is closed
A ∪ B is closed, so by (??) we have A ∪ B ⊆ A ∪ B = A ∪ B. To prove the
opposite inclusion fix x ∈ A ∪ B. Then x ∈ A or x ∈ B. Hence B(x, r) ∩ A 6= ∅ or
B(x, r) ∩ B 6= ∅ for all r > 0, that is, B(x, r) ∩ (A ∪ B) 6= ∅ for all r > 0. Hence
x ∈ A ∪ B, showing that A ∪ B ⊆ A ∪ B. Combining the two inclusions we get
A ∪ B = A ∪ B.
(c) Show that A ∩ B ⊆ A ∩ B with possibly proper inclusion.
Solution: Note that A ∩ B ⊆ A ∩ B. An intersection of two closed sets is closed.
Hence A ∩ B is closed, and by (??) we have A ∩ B ⊆ A ∩ B = A ∩ B. To see that
there is no equality in general consider A = (0, 1) and B := (1, 2) Then A ∩ B = ∅,
so A ∩ B = ∅. On the other hand A = [0, 1] and B := [1, 2], so A ∩ B = {1}.
Clearly {1} 6⊆ ∅.
Challenge questions (optional)
*8. Denote by GLN (K) the set of invertible matrices in KN ×N . We proved in Tutorial 6,
Question 7 that I − B is invertible if kBk < 1. Use this to prove that GLN (K) is open
in KN ×N .
(Note: The above is a way to see that if we perturb the coefficients of an invertible matrix
slightly, it will stay invertible. The proof does not make use of determinants, and the
ideas apply to more general situations.)
Solution: Let A be any invertible matrix in KN ×N . We have to show that there
exists r > 0 such that the open ball about A with radius r is contained in GLN (K). In
other words, there exists r > 0 such that every matrix B in KN ×N is invertible when
kA − Bk < r. We write
B = A I − (I − A−1 B) .
By Tutorial 6, Question 7, the matrix B is invertible if
kI − A−1 Bk < 1.
(1)
We have
kI − A−1 Bk = kA−1 (A − B)k ≤ kA−1 kkA − Bk
so (??) is satisfied if
kA − Bk < r :=
1
.
kA−1 k
Hence, it follows that every matrix B in KN ×N is invertible if kA − Bk < r. This proves
that GLN (K) is open in KN ×N .
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