Show that any finite subset of Rn is closed

1. hw 5

Recall: Let S be a subset of R. We say that S is bounded below if there exists a ∈ R such that x ≥ a for any x ∈ S. In this case, a is called a lower bound for S. Assume that S is bounded below.

We say that a lower bound L for S is the greatest lower bound for S if a is a lower bound for S, then a ≤ L. The greatest lower bound for S is denoted by inf S.

(1) A subset S of Rⁿ is finite if number of elements of S are finite. Show that any finite subset of Rⁿ is closed.

Proof. Let p be a point of Rⁿ. Let us show that {p} is a closed subset of Rⁿ. It is equivalent to show that Rⁿ\ {p} is open. For q ∈ Rⁿ\ {p}, choose  = d(p, q)/2. We leave it to the reader to verify that B(q, ) is contained in Rⁿ\{p}. (We have done similar examples before.) We show that Rⁿ\ {p} is open.

Let S be a finite subset of Rⁿ. Then S = {p1, · · · , pk}. Since any finite union of closed sets is closed, S =Sk

i=1{pk} is closed.

 (2) Let A be a finite subset of Rⁿ and

B = {x ∈ Rⁿ: d(x, y) ≤ 1 for some y ∈ A}.

Show that B is closed.

Proof. For any y ∈ A, show that the set D(y, 1) = {x ∈ Rⁿ : d(x, y) ≤ 1} is closed. Since A is a finite subset of Rⁿ and any finite union of closed set is closed,

B = [

y∈A

D(y, 1)

is closed. 

(3) Let A be a subset of Rⁿ and A⁰ be its derived set (the set of all accumulation points) of A.

Prove that A⁰ is closed. Is (A⁰)⁰= A⁰ for all A?

Proof. Let A = {1/n : n ∈ N} be a set considered as a subset of R. The set of accumulation point of A consists of a single point A⁰ = {0}. We see that A⁰ consists of an isolated point of A⁰. Thus (A⁰)⁰ = ∅ 6= A⁰.

 (4) Let S = {(r, s) ∈ R²: r, s ∈ Q}.

(a) Find S.

S = R².

(b) Find S^c. Here S^c= R²\ S.

S^c= R². (c) Find ∂S.

Using the previous results, we find ∂S = R².

(5) We have two equivalent definitions of convergence of sequences in R^p.

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Definition 1.1. (Defn I) A sequence (an) in R^p is convergent to a point a ∈ R^p if for any

 > 0, there exists a natural number N (this natural number depends on the choice of ) so that

an∈ B(a, ) whenever n ≥ N.

Definition 1.2. (Defn II) A sequence (a_n) in R^pis convergent to a point a ∈ R^pif for any neighborhood U of a, there exists a natural number NU of a such that

a_n∈ U whenever n ≥ NU.

We have proved in class that Defn I is equivalent to Defn II. Students may not be familiar with Defn II. In this exercise, we will let you experience how NU depends on the choice of U.

Let (an) be the sequence defined by an=

 sin1

n,1 n



, n ∈ N.

(a) Use Defn I to show that (an) is convergent to 0 = (0, 0).

Proof. By 0 < sin(1/n) < 1/n for any n ≥ 1, kank =

r sin² 1

n+ 1 n² <

r2 n² < 2

n.

For any  > 0, we choose N_ = [2/] + 1. Then for any n ≥ N_, ka_nk < . In other words, a_n∈ B(0, ) whenever n ≥ N. Here 0 = (0, 0).  (b) Let A = {(x, y) : x²+ 4y²≤ 10⁻⁵}. Show that A is a neighborhood of 0, i.e. find  > 0 so that B(0, ) ⊂ A. (A is a neighborhood of 0 if and only if 0 is an interior point of A.) Use this  to find a natural number N so that an∈ A for all n ≥ N. This number will denoted by NA.

Proof. The key point is this: 0 is an interior point of A. Find an  such that B(0, ) ⊂ A. For example, you might choose  = 1/10⁴. Assuming (x, y) ∈ B(0, 1/10⁴). Then

|x| < 1/10⁴ and |y| < 1/10⁴ and hence x²+ 4y²< 5

10⁸ < 1 10⁵.

This implies B(0, 1/10⁴) ⊂ A. Using the result in (a) and  = 1/10⁴, we take N_ = [2 · 10⁴] + 1 = 2 · 10⁴+ 1. For any n ≥ 2 · 10⁴+ 1, one has

an∈ B(0, 1/10⁴) ⊂ A.

Hence an ∈ A whenever n ≥ 2 · 10⁴+ 1. We take NA= 2 · 10⁴+ 1.

 (c) Let B =



(x, y) : − 1

10⁶ ≤ x, y ≤ 1 10⁶



. Show that B is a neighborhood of 0 i.e. find δ > 0 so that B(0, δ) ⊂ B. Use δ to find a natural number M so that an ∈ B for all n ≥ M. This number will denoted by M = N_B.

Proof. We simply take  = 1/10⁶. For (x, y) ∈ B(0, 1/10⁶), we have |x|, |y| < 1/10⁶. Thus B(0, 1/10⁶) ⊂ B. Using the result in (a), we can choose N= [2·10⁶]+1 = 2·10⁶+1.

Then

an ∈ B(0, 1/10⁶) ⊂ B

whenever n ≥ 2 · 10⁶+ 1. This implies that an∈ B whenever n ≥ 2 · 10⁶+ 1. We thus take NB= 2 · 10⁶+ 1.



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(6) Prove that the following sequences have a convergent subsequence in the given spaces.

We use the Bolzano-Weierstrass Theorem to prove the results. We show that these se- quences are bounded.

(a) Let (an) be the sequence of real numbers in R defined by an= sin n, n ∈ N.

Proof. Since |an| ≤ 1 for any n ≥ 1, (an) is bounded.

 (b) Let (bn) be the sequence of real numbers in R defined by

bn= (−1)ⁿsin eⁿ, n ∈ N.

Proof. We know |bn| = | sin(eⁿ)| ≤ 1 for any n ≥ 1. Hence (bn) is bounded.

 (c) Let (a_n) be the sequence in R²defined by

an = (cos n, sin n), n ∈ N.

Proof. Since kank = 1 for all n ≥ 1, (an) is bounded.

 (d) Let (b_n) be the sequence in R³ defined by

bn=



cos n, e⁻ⁿcos n², cos1 n



, n ∈ N.

Proof. Since | cos n| ≤ 1 and |e⁻ⁿcos n²| ≤ e⁻¹ and | cos(1/n)| ≤ 1 for all n ≥ 1, (b_n) is bounded.