1. Quizz 2 True or false.
(1) ( ) The origin (0, 0) is an accumulation point of U = {(x, y) ∈ R2 : x > 0}.
Proof. The statement is true.
Denote (0, 0) by p. For every > 0, we choose q= (/2, 0). Since > 0, /2 > 0.
The point q∈ U. On the other hand, d(p, q) =
2 < .
Hence q∈ B(p, ). We conclude that q ∈ B(p, ) ∩ U. Thus B(p, ) ∩ U is nonempty for any > 0. By definition, p is an accumulation point of U. (2) Let A = {(m, 0) ∈ R2 : m ∈ Z}.
(a) ( ) A consists of isolated points only.
Proof. The statement is true.
Let p ∈ A. Then p = (m, 0) for some integer m. Choose = 1/2. Let us show that
B(p, ) ∩ A = {p}.
By definition of p, p ∈ B(p, ) ∩ A. Thus {p} ⊂ B(p, ) ∩ A.
Let q ∈ B(p, ) ∩ A. Then q ∈ A and q ∈ B(p, ). Since q ∈ A, q = (n, 0) for some integer n. Since q ∈ B(p, ), d(p, q) < 1/2. Since
d(p, q) =p
(m − n)2+ 02 = |m − n|,
we find |m − n| < 1/2. Since both m, n are integers, m = n. This implies that p = q. Hence B(p, )∩A ⊂ {p}. We conclude that B(p, )∩A = {p}. By definition, p is an isolated point of A. We prove that every point of A is an isolated point of A.
(b) ( ) The set int(A) is nonempty.
Proof. The statement is false.
The set int(A) is empty. The proof is given as follows.
Suppose int(A) is nonempty. Let p be an interior point of A. Then we can choose > 0 so that B(p, ) ⊂ A. We can choose so that < 1. Then B(p, ) = B(p, ) ∩ A = {p}. Since {p} is contained in B(p, ), the previous result shows that B(p, ) = {p} which is impossible. Thus no such p exists; int(A) must be empty.
(c) ( ) Let A0 be the set of all accumulation points of A. Then A0 = ∅.
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Proof. The statement is true.
Let us show that all points of Rn are not accumulation points of A and hence A0 = ∅.
If p ∈ A, then p is an isolated point of A. Since an isolated point of A is never an accumulation point of A, we find that p is not an accumulation point of A when p ∈ A.
Let p = (x0, y0) ∈ Rn\ A. Let us study the case when y0 = 0. Since p 6∈ A, x0
is not an integer. We choose n ∈ Z so that n ≤ x0 < n + 1. (n = [x0]) Choose
> 0 such that < min{x0− n, n + 1 − x0}. Claim that B(p, ) ∩ A = ∅. Suppose not. Choose q ∈ B(p, ) ∩ A. Write q = (m, 0) for some integer m. Then
d(p, q) = |x0− m| < .
Since < x0 − n, |x0− m| < x0 − n. We find m > n. Since < n + 1 − x0,
|x0− m| < n + 1 − x0. We find m < n + 1. We conclude that n < m < n + 1.
Since both m, n are integers, the inequality n < m < n + 1 never happens.
Let us prove the case when y0 6= 0. In this case, we choose = |y0|/2 > 0. Claim B(p, ) ∩ A = ∅. Suppose not. We choose q ∈ B(p, ) ∩ A. Write q = (m, 0) for some m ∈ Z. Then
d(p, q) = q
(m − x0)2+ y20 < |y0| 2 . This implies that (m − x0)2+ y20 < y02/4 and hence
(m − x0)2+3 4y02 < 0 which is impossible.
(d) ( ) A is a closed subset of R2.
Proof. The statement is true.
Since A0= ∅, A = A. Thus A is closed by definition. (3) Let D = {x2+ y2 < 1} and p = (0, 1) and O = (0, 0).
(a) ( ) p is an adherent point of D.
Proof. The statement is true. Exercise.
(b) ( ) p is an exterior point of D.
Proof. The statement is false. p is a boundary point of D and thus any -ball
center at p intersects with Dc.
(c) ( ) p is an isolated point of D.
Proof. The statement is false. It is an accumulation point of D. (d) ( ) p is a boundary point of D.
Proof. The statement is true. The proof is already given in the example text.
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(e) ( ) O is an interior point of D.
Proof. The statement is true. You can choose = 1/2. Then B(O, 1/2) ⊂ D. (f) ( ) O is an adherent point of D.
Proof. The statement is true. See below.
(g) ( ) O is an accumulation point of D.
Proof. The statement is true. For any > 0, choose a real number x so that 0 < x< when < 1 and choose x= 1/2 when > 1. Take p = (x, 0). Then p∈ B0(O, ) ∩ D. Thus O is an accumulation point of D. (h) ( ) The boundary of D is {x2+ y2 = 1}.
Proof. The statement is true. The proof is given in the example. (4) Let S = {(x, y) ∈ R2 : 0 ≤ x ≤ 1, 3 ≤ x ≤ 5}.
(a) ( ) int(S) = {(x, y) ∈ R2 : 0 < x < 1, 3 < x < 5}.
Proof. The statement is true. Proof is left to the readers as an exercise.
(b) ( ) ∂S = {(x, y) ∈ R2: x = 0, x = 1, x = 3, x = 5}.
Proof. The statement is true. Proof is left to the readers as an exercise. (5) ( ) An isolated point of a subset of Rn belongs to the set.
Proof. The statement is true. It is proved in class.
(6) ( ) An isolated point of a subset of Rn can be an interior point of that set.
Proof. The statement is false. An isolated point of a subset of Rn can never be an interior point of that set.
(7) ( ) An accumulation point of a subset of Rn could be the interior point of that
set.
Proof. The statement is true. Consider the open ball B = B(p, ). Its center is an interior point of B while it is also an accumulation point of B.
(8) ( ) An accumulation point of a subset of Rnmust be the interior point of that
set.
Proof. The statement is false. Take p = (0, 1) and S = {(x, y) : x2+ y2 < 1}. Then p is an accumulation point of S but p is not an interior point of S.
(9) ( ) An accumulation point of a subset of Rn can never be the exterior point of
that set.
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Proof. The statement is true. Let p be an accumulation point of S. If p ∈ ext(S), then there exists > 0 so that B(p, ) ⊂ Sc. Thus B(p, ) ∩ S = ∅. This shows that p is not an accumulation point of S which leads to a contradiction.
