Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2014
Exam #2
Solutions
This is a take-home examination. The exam includes 6 questions. The total mark is 100
points. Please show all the work, not only the answers.
1. [17 points] Solve Exercise 92 in Chapter 2 of the textbook.
Solution:
(a) We have (p. 66 in the textbook): ∂S = S ∩ S c . Thus ∂S = ∅ if and only if
S ∩ S c = ∅.
Since S = lim S (Proposition 12, p. 67 in the textbook), ∂S = ∅ if and only if
lim S ∩ lim S c = ∅.
(1)
Since S c ⊂ lim S c and S ⊂ lim S, (1) implies
lim S ∩ S c = ∅
and
S ∩ lim S c = ∅.
The first identity says that S is closed while second one tells us that S c is closed. Thus
if ∂S = ∅, then S is both closed and open (as the complement to a closed set S c ).
On the other hand, if S is clopen then both S and S c are closed and hence S = S
while S c = S c , implying that
S ∩ S c = S ∩ S c = ∅.
(b) The claim follows immediately from the formula ∂S = S∩S c and the identity (S c )c = S.
(c) Since both S and S c are closed, ∂S = S ∩ S c is a closed set. For any closed set F we
have
∂F = F ∩ F c ⊂ F = F.
Since F = ∂S is closed, ∂∂S ⊂ ∂S.
(d) Then result follows from the following lemma applied to A = ∂S. Notice that ∂S is a
closed set because it is the intersection of the closed sets S and S c .
Lemma. If A is a closed set in a metric space, then ∂∂A = ∂A.
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Proof of the lemma. We know from the result in (c) that ∂∂A ⊂ ∂A. Furthermore,
according to the result in part (a) of Exercise 91 (see solutions to HW 3),
int B = B\∂B,
and hence, setting B = ∂A,
∂∂A = ∂A ⇔ ∂B = B ⇔ int B = ∅.
Thus it suffices to show that
int B = ∅,
where B = ∂A.
(2)
To get a contradiction, assume that int B 6= ∅. Pick any q ∈ int B. Then
p ∈ B = ∂A ⊂ Ac ⊂ Ac .
Since Ac is open, q is its interior point. Therefore, there exists r > 0 such that
Mr (q) ⊂ Ac .
In particular, Mr (q) ∩ A = ∅, and hence
q 6∈ lim A = A.
But ∂A ⊂ A = A, and hence the latter assertion implies that q 6∈ ∂A = B, yielding
the contradiction to the assumption that q ∈ int B. This completes the proof of (2)
and hence of the lemma.
(e) Using the fact that S = int S ∪ ∂S, the result can be quickly derived from the result
in Exercise 91(c) (se HW 3). Here is a direct proof:
We have
∂(S ∪ T ) = S ∪ T ∩ (S ∪ T )c = S ∪ T ∩ S c ∩ T c
(3)
Observe that
• p ∈ S ∪ T = lim S ∪ T means that there is a sequence pn S ∪ T converging to p.
The sequence has infinitely many members in at least one of the sets S and T,
forming a subsequence converging to p and contained entirely in one of the sets.
Thus p is a limit point of either S or T. In other words,
S ∪ T ⊂ S ∪ T.
(4)
• Similarly, p ∈ S c ∩ T c = lim(S c ∩ T c ) means that there is a sequence pn ∈ S c ∩ T c
converging to p. Since pn ∈ S c ∩ T c implies pn ∈ S c and pn ∈ T c , p is a limit point
of both S c and T c . In other words,
S c ∩ T c ⊂ S c ∩ T c.
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(5)
It follows from (3), (4), and (5) that
∂(S ∪ T ) ⊂ S ∩ S c ∪ T ∩ T c = ∂S ∪ ∂T,
as required.
(f) For instance, if M = R equipped with the usual Euclidean metric and S = Q, then
∂S = ∂Q = R while
∂∂S = ∂R = R ∩ Rc = R ∩ ∅ = R ∩ ∅ = ∅.
(g) Similarly to (f), let M be R equipped with the usual Euclidean metric, S = Q, and
T = Qc . Then S ∪ T = R, and hence
∂(S ∪ T ) = ∂R = ∅,
while ∂S = ∂T = R, and hence
∂S ∪ ∂T = R ∪ R = R.
2. [17 points]
(a) Solve Exercise 11 in Chapter 2 of the textbook.
(a) Solve Exercise 12 in Chapter 2 of the textbook.
Solution:
(a) A natural example of a bijection f : T → K is given by f (A) = Ac .
(b) Let (M0 , d0 ) be a metric space with the discrete metric and (M, d) be a metric space
homeomorphic to (M0 , d0 ). Let h : M → M0 be a homeomorphism between two spaces.
– Solution to 12(a):
Let E be an arbitrary subset of M0 . If x ∈ E then Mr (x) = {x} ⊂ E for any
r ≤ 1. Hence E is open. Now consider an arbitrary subset A of M. Since h is a
bijection, A = h−1 (E) for some E ∈ M0 . Since h is continuous and E is open in
M0 , it follows that A is open in M. Furthermore, since Ac is open by the above
argument, A is also closed. Thus any subset A of M is clopen.
– Solution to 12(b):
Let (N, dN ) be a metric space and consider a function f : M → N. Since any
subset of M is open, f −1 (E) is open for any E ⊂ N. Thus f is continuous by
Theorem 10(iii) in Chapter 2 of the textbook.
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– Solution to 12(c): Let xn ∈ M, n ∈ N, be an arbitrary sequence of elements
of M. Assume that xn converges in M and x = limn→∞ xn . Then, since the
homeomorphism h is continuous, h(x) = limn→∞ h(xn ) in M0 . But the distance
between any two distinct points of M0 equals one according to the definition of the
discrete metric. Therefore, the convergence in M0 is only possible if h(xn ) = h(x)
for all n large enough. Since h is a bijection, this implies that xn = x for all n
large enough. In other words, the sequence xn converges in M if and only if there
exist n0 ∈ N and x ∈ M such that xn = x for all n ≥ n0 .
3. [17 points]
(a) Solve Exercise 78(b) in Chapter 2 of the textbook.
(b) Solve Exercise 78(d) in Chapter 2 of the textbook.
Solution:
(a) Suppose that M is a compact and chain-connected metric space. If M is disconnected,
then M = A ∪ Ac for some proper clopen set A in M. Then, by Theorem 35 in the
textbook, both A and Ac are compact. It follows from the result in Exercise 71 of
Chapter 2 (see Problem 3(d) in the practice test) that
s :=
inf
x∈A,y∈Ac
d(x, y) > 0.
Therefore, there is no ε-chain in M for any ε < 21 s. This is in contrary to the assumption
that M is chain-connected. Thus M must be connected.
(b) M doesn’t have to be connected even if it is complete and chain-connected. For instance, consider the union of the curves y = x and y = x + e−x , x ∈ R, as a subset of
R2 . Both the curves are closed subsets of a complete space R2 , and hence their union
is a complete set (Theorem 27 on p.74 of the textbook). Furthermore, the union is
chain-connected because limx→∞ e−x = 0 and hence it is possible to “hop” from one of
the curves to another by a jump of arbitrarily small size.
4. [16 points] Sets A and B in a metric space are called separated if A∩B = ∅ and A∩B = ∅.
(a) If A and B are two disjoint closed sets in some metric space X, prove that they are
separated.
(b) Prove the same for disjoint open sets.
(c) Fix p ∈ X, δ > 0, and define
A = {q ∈ X : d(p, q) < δ}
and
Prove that A and B are separated.
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B = {q ∈ X : d(p, q) > δ}.
(d) Use the result in (c), to show that every connected metric space with at least two
points is uncountable (the result is Exercise 56 in Chapter 2).
Solution:
(a) Since the sets are closed, A = A and B = B. Hence, if x ∈ A ∩ B or x ∈ A ∩ B, then
actually x ∈ A ∩ B. But A ∩ B is assumed to be empty.
(b) Suppose that x ∈ A ∩ B. Since A ∩ B = ∅, then x ∈ A ∩ B 0 , where B 0 denotes the
set of cluster points of B. Since A is open, Mr (x) ⊂ A for some r > 0. Here, as usual,
Mr (x) denotes the open ball of radius r centered at x. However, if x is a cluster point
of B, then A ∩ B is not empty. Contradiction.
(c) These are two open disjoint sets. They are separated by the result in (b).
(d) Take any two distinct points p, p1 ∈ X. Pick any real constant δ > 0 such that
δ < d(p, p1 ). Then the result in (c) along with the definition of connected sets show
that there exists xδ ∈ X such that d(p, xδ ) = δ. Indeed, if no such xδ exists for some δ
between 0 and d(p, p1 ), then the sets A and B defined in (b) are clopen and their union
is covering X. Since δ varies within an interval of reals I = {x ∈ R : 0 ≤ x ≤ d(p, q)}
and δ → xδ is an injection from I onto X, the set X is uncountable.
5. [16 points] Solve Exercise 63 in Chapter 2 of the textbook. Use the following definition
of the compact sets in R2 : a set A ⊂ R2 is called compact if it is closed and bounded.
Solution: In crude terms, the key technical observation used in the solution is as follows
(see more details below). The set A is a starlike set with a convex boundary. Let the origin
O be a center of A and consider any point K in the boundary. If the polar coordinates of
K are K = (Rθ , θ) and you can find two points M = (r1 , θ1 ) ∈ A and N = (r2 , θ2 ) ∈ A with
θ1 < θ < θ2 but θ1 ≈ θ and θ2 ≈ θ, then the four segments OM, ON, KM, and KN are
contained within A. Thus the whole quadruple OM KN is a subset of A and hence all points
in the open segment connecting OK are interior points of A. Furthermore, essentially the
same argument shows that Rθ is a continuous function of θ. Indeed, boundary points with
the angular coordinate close to θ are within the cone formed by the arrays OM and ON, and
while they must stay outside of the polygon OM KN they cannot cluster anywhere above
K in the array OK.
(a) Let A ⊂ R2 be a convex, closed set with non-empty interior. We must prove
A = int A.
To understand the geometric structure of A we will use polar coordinates (r, θ) in R2 .
We will denote by O the origin in R2 . Exercise 61(a) claims that A is a starlike set
(such sets are also considered in Exercise 62 which is a part of the practice exam).
Observe that if P1 = (r1 , θ1 ) ∈ A and P2 = (θ2 , R2 ) ∈ A, the set A contains the
segment connecting P1 and P2 , and hence for any θ between θ1 and θ2 there is a point
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with angular coordinate θ that belongs to A. Assuming (without a loss of generality)
that A intersect the axis 0x at least in two different pints, and taking in account that
A is a closed set, we obtain that in polar coordinates A has the following form:
A = {(θ, r) : 0 ≤ θ ≤ θA , 0 ≤ r ≤ Rθ }
(6)
for some constant θ ∈ (0, 2π] and a function Rθ ∈ (0, +∞]. Notice that Rθ can take
the value +∞ and that θA > 0 since the interior of A is assumed to be non-empty.
Thus the boundary consists of points with r = Rθ for which Rθ < +∞ and, possibly,
of points with θ ∈ {0, θA }. Let
ΩA = (θ, r) ∈ A : either θ ∈ {0, θA } or r = Rθ < ∞ .
Thus ∂A ⊂ ΩA . Notice that any point in ΩA can be connected to a point in A\ΩA by
a strict segment included entirely in A\ΩA , and hence is a limit point of a sequence of
points belonging to this segment.
To complete the proof it remains to verify that
A\ΩA ⊂ int A.
To this end, observe that if P ∈ A\ΩA , then we can find two straight segments, say
KL and M N, contained entirely within the set A, such that P is the intersection of
this two segments. Thus (draw a picture of two segments and their intersection) P lies
within a convex polygon KM LN contained in the set A. Since P is an interior point
of the polygon, it is also an interior point of A.
(b) We will need the following lemma.
Lemma. Suppose that A is closed, bounded, and convex subset of R2 with a non-empty
interior. Suppose (without loss of generality) that A intersects the axis 0x in at least
two different points, and hence the representation (6) is valid. Then, the function Rθ
introduced in (6) is a continuous function of the parameter θ in (0, θA ).
Proof of the lemma. Notice that Rθ < ∞ for all θ since A is bounded. Assume that
Rθ is not continuous at some θ∗ ∈ (0, θA ). Then there exists ε > 0 and θn such that
lim θn = θ∗
n→∞
but |Rθn − Rθ∗ | > ε.
Since A is a closed bounded (and hence compact) set in R, there is a subsequence
(θnk , Rθnk ) of (θn , Rθn ) converging to a point (θ∗ , R∞ ) ∈ A. Moreover,
|Rθ∗ − R∞ | ≥ ε.
Due to the definition of Rθ , we must have R∞ ≤ Rθ∗ . Hence
0 ≤ Rθ∗ ≤ R∞ − ε.
Now, consider two pints (θ1 , r1 ) and (θ2 , r2 ) with θ1 < θ∗ < θ2 and θ1 , θ2 close enough
to θ, so that the polygon with 4 edges and the vertices located in the origin, (θ∗ , Rθ∗ ),
(θ1 , r1 ) and (θ2 , r2 ) is convex and contains (θ∗ , R∞ ). The point (θ∗ , R∞ ) is an interior
point of the polygon (and hence of A), and therefore it cannot be a limit point of the
boundary. The contradiction shows that Rθ is a continuous function of θ.
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Write a representation for B similar to the one in (6):
B = {(θ, r) : 0 ≤ θ ≤ θB , 0 ≤ r ≤ Sθ }.
By moving if necessary the origin of the coordinate system to inner points of A and
B, we can assume without loss of generality that θA = θB = 2π. Use the above lemma
to conclude that the function f : A → B defined by
f (r, θ) =
S
θ
Rθ
r, θ
is a homeomorphism between A and B.
6. [17 points] Solve Exercise 13 (Pre-lim problems in Chapter 2) on p. 136 of the textbook.
Solution:
(a) Assume the contrary, namely the sequence xn is divergent. Since A is a compact set,
there is a converging subsequence xnk of xn . By virtue of the definition of the point x,
lim xnk = x.
k→∞
Since xn is divergent, there exists ε > 0 with the following property. For any j ∈ N
there exists nj > j such that
d(xnj , x) > ε.
(7)
Let yj = xnj , j ∈ N. Clearly, in view of (7), no subsequence of the sequence yj can
converge to x. But since A is compact and due to the definition of x, such a subsequence
must exists. The contradictions shows xn is convergent. Thus limn→∞ xn = x.
(b) For instance, consider (xn )n≥0 such that x0 = 0, and x2k−1 = 1/k while x2k = k for
k ∈ N. The only cluster point of this sequence is x0 = 0, but the sequence is unbounded
and hence divergent.
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