Chapter 7.3, Problem 3E

a.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{\frac{n+1}{2n}:n\in \mathbb{N}\right\}$

Calculation:

Consider the set $A=\left\{\frac{n+1}{2n}:n\in \mathbb{N}\right\}$

  $A=\left\{\frac{1}{2}+\frac{1}{2n}:n\in \mathbb{N}\right\}$

Here $\frac{1}{2n}$ converges towards $0$ as $n\in \mathbb{N}$

Therefore the derived set of $A$ is $A\text{'}=\left\{\frac{1}{2}\right\}$

b.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{2^n:n\in \mathbb{N}\right\}$

Calculation:

Consider the set $B=\left\{2^n:n\in \mathbb{N}\right\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in B$

Now, ${\rm N}(x,\delta )-x\cap B=\varphi$

Therefore the derived set of $B$ is $B\text{'}=\varphi$

c.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{6n:n\in \mathbb{N}\right\}$

Calculation:

Consider the set $A=\left\{6n:n\in \mathbb{N}\right\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-x\cap A=\varphi$

Therefore the derived set of $A$ is $A\text{'}=\varphi$

d.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{\frac{7}{2n}:n\in \mathbb{N}\right\}$

Calculation:

Consider the set $A=\left\{\frac{7}{2n}:n\in \mathbb{N}\right\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Here $\frac{7}{2n}$ converges towards $0$ as $n\in \mathbb{N}$

Now, ${\rm N}(x,\delta )-\left\{0\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=\left\{0\right\}$

e.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $(0,1]$

Calculation:

Consider the set $A=(0,1]$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=[0,1]$

f.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $(3,7)\cup \{2,6,8\}$

Calculation:

Consider the set $A=(3,7)\cup \{2,6,8\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Then ${\rm N}(x,\delta )$ contains infinite points from the open interval $(3,7)$ but no points from the set $\{2,6,8\}$ other than $2,6and8$ .

Therefore the derived set of $A$ is $A\text{'}=[3,7]$

g.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{1+\frac{{(-1)}^{n}n}{n+1}:n\in \mathbb{N}\right\}$

Calculation:

Consider the set $A=\left\{1+\frac{{(-1)}^{n}n}{n+1}:n\in \mathbb{N}\right\}$

  $\begin{array}{l}A=\left\{1+\frac{{(-1)}^{n}n}{n+1}:n\in \mathbb{N}\right\}\\ =\left\{1-\frac{1}{2},1+\frac{2}{3},1-\frac{3}{4},1+\frac{4}{5},\mathrm{....}\right\}\\ =\left\{\frac{1}{2},\frac{5}{3},\frac{1}{4},\frac{9}{5},\mathrm{....}\right\}\\ =\left\{\frac{1}{2},\frac{1}{4},\mathrm{....}\right\}\cup \left\{\frac{5}{3},\frac{9}{5},\mathrm{...}\right\}\end{array}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=\{0,2\}$

h.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\mathbb{Q}\cap (0,1)$

Calculation:

Consider the set $A=\mathbb{Q}\cap (0,1)$

  $A=\left\{x\in \mathbb{Q}:0<x<1\right\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=[0,1]$

i.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{\frac{1+n^2(1+{(-1)}^n}{n}:n\in \mathbb{N}\right\}$

Calculation:

Consider the set $A=\left\{\frac{1+n^2(1+{(-1)}^n}{n}:n\in \mathbb{N}\right\}$

  $\begin{array}{l}A=\left\{\frac{1+n^2(1+{(-1)}^n}{n}:n\in \mathbb{N}\right\}\\ A=\left\{\frac{1}{n}+n\left(1+{(-1)}^n\right):n\in \mathbb{N}\right\}\end{array}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=\left\{0\right\}$

j.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{\sin x:x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\right\}$

Calculation:

Consider the set $A=\left\{\sin x:x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\right\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=[-1,1]$

k.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{k+\frac{1}{n}:k,n\in \mathbb{N}\right\}$

Calculation:

Consider the set $A=\left\{k+\frac{1}{n}:k,n\in \mathbb{N}\right\}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=\{k:k\in \mathbb{N}\}$

l.

To determine

To find the derived set of the given set.

Explanation of Solution

Given:

  $\left\{\frac{x}{2^y}:x,y\in \mathbb{Z}\right\}$

Calculation:

Consider the set $A=\left\{\frac{x}{2^y}:x,y\in \mathbb{Z}\right\}$

  $\begin{array}{l}A=\left\{\frac{x}{2^y}:x,y\in \mathbb{Z}\right\}\\ A=\left\{\frac{x}{2^y}:x\in \mathbb{Z},y\in {\mathbb{Z}}^{+}\right\}\cup \left\{\frac{x}{2^y}:x\in \mathbb{Z},y\in {\mathbb{Z}}^{-}\right\}\end{array}$

Suppose ${\rm N}(x,\delta )$ be a neighborhood point at any point $x\in A$

Now, ${\rm N}(x,\delta )-\left\{x\right\}\cap A\ne \varphi$

Therefore the derived set of $A$ is $A\text{'}=\left\{0\right\}$