Chapter 5.3, Problem 9E

a.

To determine

Prove that an infinite subset of a denumerable set is denumerable.

Answer to Problem 9E

Proved

Explanation of Solution

Given information:

Use the theorems of this section.

Calculation:

We have to use the theorem to prove an infinite subset of a denumerable set is denumerable.

A denumerable set $K$ and assume $H$ is an infinite subset of $K$ .a finite or a denumerable set is countable since set $K$ is denumerable so it is countable.

Every subset of a countable set is countable so $H$ is countable.

Since $H$ is an infinite subset of $K$ , $H$ is denumerable.

Hence, an infinite subset of a denumerable set is denumerable.

b.

To determine

Prove that if $A$ is a countable subset of an uncountable set $B$ , then $B-A$ is uncountable.

Answer to Problem 9E

Proved

Explanation of Solution

Given information:

Use the theorems of this section.

Calculation:

We have to use the theorem to prove $A$ is a countable subset of an uncountable set $B$ , then $B-A$ is uncountable.

Consider an uncountable set $B$ as $B=\left(B\backslash A\right)\cup A$ . Assume that the set $B\backslash A$ is countable. Since $A$ is countable and countable union of countable set is countable, so set $B=\left(B\backslash A\right)\cup A$ is also countable.

It implies that set $B$ is countable. But $B$ is uncountable set.

The assumption $B\backslash A$ is incorrect.

Hence, if $A$ is a countable subset of an uncountable set $B$ , then $B-A$ is uncountable.

c.

To determine

Prove that $Q\cap \left(1,2\right)$ is denumerable.

Answer to Problem 9E

Proved

Explanation of Solution

Given information:

Use the theorems of this section.

Calculation:

We have to use the theorem to prove $Q\cap \left(1,2\right)$ is denumerable.

The set $Q\cap \left(1,2\right)$ is the collection of all the rational numbers that exist in the interval $\left(0,1\right)$ . Since the set $Q$ of all rational numbers is denumerable , so the set of the rational numbers between the interval $\left(0,1\right)$ is also denumerable.

Hence, $Q\cap \left(1,2\right)$ is denumerable.

d.

To determine

Prove that $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is denumerable.

Answer to Problem 9E

Proved

Explanation of Solution

Given information:

Use the theorems of this section.

Calculation:

We have to use the theorem to prove $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is denumerable.

The set $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is same as,

  $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)=\left(Q\cap \left(1,2\right)\right)\cup \left(Q\cap \left(2,3\right)\right)\cup \left(Q\cap \left(3,4\right)\right)\cup \mathrm{..}\cup \left(Q\cap \left(19,20\right)\right)$

The set $\left(Q\cap \left(1,2\right)\right)$ is the collection of all rational numbers that comes in the interval $\left(0,1\right)$

Since the set $Q$ of all rational numbers is denumerable , so the set of the rational numbers between the interval $\left(0,1\right)$ is also denumerable.

Hence, $\left(Q\cap \left(1,2\right)\right)$ is denumerable. Similarly the sets

  $\left(Q\cap \left(2,3\right)\right)\cup \left(Q\cap \left(3,4\right)\right)\cup \mathrm{..}\cup \left(Q\cap \left(19,20\right)\right)$ are denumerable. Since

  $\left(Q\cap \left(1,2\right)\right)\cup \left(Q\cap \left(2,3\right)\right)\cup \left(Q\cap \left(3,4\right)\right)\cup \mathrm{..}\cup \left(Q\cap \left(19,20\right)\right)$ is a disjoint family of denumerable sets, therefore, the set $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is countable.

The set $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is either denumerable or finite.

Hence, $\underset{n=1}{\overset{20}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is denumerable.

e.

To determine

Prove that $\underset{n\in N}{\overset{n=1}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is denumerable.

Answer to Problem 9E

Proved

Explanation of Solution

Given information:

Use the theorems of this section.

Calculation:

We have to use the theorem to prove $\underset{n\in N}{\overset{n=1}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is denumerable.

The set $\left(Q\cap \left(n,n+1\right)\right)$ for some $n\in Q$ is denumerable. As the countable collection of countable seta is countable, the set $\underset{n\in N}{\overset{n=1}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is countable. The set $\underset{n\in N}{\overset{n=1}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is infinite.

Hence, $\underset{n\in N}{\overset{n=1}{{\displaystyle \cup }}}\left(Q\cap \left(n,n+1\right)\right)$ is denumerable.

f.

To determine

Prove that $\underset{n\in N}{\overset{}{{\displaystyle \cup }}}\left\{\frac{n}{2^k}:k\in N\right\}$ is denumerable.

Answer to Problem 9E

Proved

Explanation of Solution

Given information:

Use the theorems of this section.

Calculation:

We have to use the theorem to prove $\underset{n\in N}{\overset{}{{\displaystyle \cup }}}\left\{\frac{n}{2^k}:k\in N\right\}$ is denumerable.

The set $\underset{n\in N}{\overset{}{{\displaystyle \cup }}}\left\{\frac{n}{2^k}:k\in N\right\}$ is in one to one correspondence with the set $\left\{\left(x,y\right):x\in N,y\in N\right\}$

Since the set of natural numbers is denumerable. the countable collection of countable sets is countable.

Hence, $\underset{n\in N}{\overset{}{{\displaystyle \cup }}}\left\{\frac{n}{2^k}:k\in N\right\}$ is denumerable.