Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Homework #4
Solutions
EXERCISES FROM CHAPTER 2
39. By Theorem 39 in the textbook (p. 79), a continuous image of a compact set is compact.
Let π : A × B → A denote the projection of A × B onto A. That is,
π(x, y) = x,
x ∈ A, y ∈ B.
It follows from part (d) of Theorem 21 in the textbook (p. 71) that π is a continuous function,
and hence A is compact whenever A × B is compact.
40.
(a) Let G denote the graph of f, that is
G = {(p, y) ∈ M × R : f p = y}.
Consider a sequence of points (pn , yn ) ∈ G converging to a point (p, y) ∈ M × R. By
the definition of G, we have yn = f (pn ). Since f is continuous,
f (p) = lim f (pn ) = lim yn = y.
n→∞
n→∞
Thus (p, y) ∈ G. This shows that G is a closed subset of M × R.
(b) If M is compact and f is continuous then f (M ) is compact. Hence, by the result in
(a), G is a closed subset of the compact set M × f (M ). By Theorem 35 in the textbook
(p. 77), a closed subset of a compact set is compact.
62.
(a) False. For instance, consider the following starlike set in R2 :
E = {(x, y) ∈ R2 : x2 + y 2 < 4}\{(x, y) ∈ R2 : y = 0 and x ≥ 1}.
(1)
The set is a disk of radius 2 centered at the origin with an excluded closed straight
segment whose endpoints are at (1, 0) and (2, 0). The segment is “shadowing” points
close to (2, 0), making some of them “invisible” while looking from a point within E.
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(b) Yes. If A ∈ Rm and B ∈ Rm are two centers, D ∈ Rm is any point in the straight
segment connecting A and B, and C ∈ E, then the segments AB, BC, and AB are
subsets of E by the definition of the center. Since any point within the triangle 4ABC
lies within a segment connecting the center point A with a point in BC which belongs
to E, then the triangle is a subset of E. In particular, the segment DC lies entirely
within E. Since C is an arbitrary point in E, this argument shows that D is a center.
Thus all points in a segment connecting any two centers are center, and hence the set
of the centers is convex.
(c) Not necessarily. For instance, consider the disk {(x, y) ∈ R2 : x2 + y 2 < 1} in R2 . Any
its point is a center, and hence the set of the centers is open.
(d) Yes. Set E introduced in (1) serves as an example.
74(a). For instance, use Corollary 50 (p. 84) in the textbook. Roman letters N and H are
connected metric spaces but their intersection consists of two disjoint (and thus disconnected)
straight vertical segments.
89. Let S 0 denote the set of the cluster points of S.
(a) If S ⊂ T and Mr (p) contains infinitely many points of S, then clearly Mr (p) contains
infinitely many points of T. This shows that any cluster point of S is a luster point of
T, and hence S 0 ⊂ T 0 .
(b) By the result in (a), since S ⊂ S ∪ T and T ⊂ S ∪ T, we have
S 0 ⊂ (S ∪ T )0
and
T 0 ⊂ (S ∪ T )0 ,
and hence
S 0 ∪ T 0 ⊂ (S ∪ T )0 .
(2)
On the other hand, if p ∈ (S ∪ T )0 then any open ball Mr (p) contains infinitely many
points of S ∪ T, and therefore it contains infinitely many points of either S or T.
Furthermore, if MR (p) contains only finitely many points of S (possible none) for some
R > 0, then the same is true for all r ∈ (0, R), which implies that Mr (p) contains
infinitely many points of T for any r ∈ (0, R). In other words, if p 6∈ S 0 then p ∈ T 0 ,
and hence p ∈ S 0 ∪ T 0 . Thus (S ∪ T )0 ⊂ S 0 ∪ T 0 . Combining this result with (2) yields
the claim.
(c) By the result in (a), since S ⊂ S,
0
S0 ⊂ S .
(3)
0
On the other hand, if p ∈ S , then any open ball Mr (p) contains infinitely many
points of S. Since the points in S are limit points of S, Mr (p) also contains infinitely
0
many points of S. Thus p ∈ S 0 . This shows that S ⊂ S 0 . Combining this result with
(3) yields the claim.
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(d) Suppose that p is a cluster point of S 0 and assume that it is not a cluster point of S.
The latter assertion means that there is an open ball Mr (p) that contains only finitely
many points of S and infinitely many points of S 0 . This is an absurd since the points
in Mr (p) ∩ S 0 must be the limit points of S and hence of Mr (p) ∩ S. However, clearly,
a finite set cannot have infinitely many limit points.
(e) N0 = ∅ and R0 = r. Furthermore, by Theorem 7 on p. 19, Q0 = R and (R\Q)0 = R.
Thus Q00 = (R)0 = R.
(f) Since limn→∞ 1/n = 0, T 0 = {0}. Thus T 00 = ({0})0 = ∅.
(g) To get an example, put S = T, where T = {1/n : n ∈ N} is the set introduced in (f).
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