**25. Show that the set of all integers greater than -100 is countable.
**

**Solution: **In order to prove that the set of all integers greater than -100 is countable, it suffices to prove that the set of integers \mathbb{Z} is countable. This follows from the fact that the subset of a countable set is countable.

We can the list the integers in a sequence

0,1,-1,2,-2,3,-3,…

We define a function

f:\mathbb{N} \rightarrow \mathbb{Z}

This is defined as

f(n)=\left\{\begin{matrix} \frac{n}{2}&\text{n is even} \\ \frac{-(n-1)}{2}& \text{n is odd}\end{matrix}\right.

We now show that f is a bijection.

First we show that it is injective by case analysis on the parity of \mathbb{N}

**Case 1**

If m,n are two even natural numbers

Then

f(m)=\frac{m}{2}

f(n)=\frac{n}{2}

It follows that

f(m)=f(n)

Implies that

m=n

**Case 2**

If m,n are two odd natural numbers

Then

f(m)=-\frac{m-1}{2}

f(n)=-\frac{n-1}{2}

It follows that

f(m)=f(n)

Implies that

m=n

Therefore f is injective.

We now show that f is surjective by case analysis on the sign of some integer t in \mathbb{Z}

**Case 1**

If t is positive, then t will appear in an even position in the sequence.

Thus,

f(2k)=\frac{2k}{2}

=t

With

t=k

This implies that for every positive value t in \mathbb{Z} there is a natural number 2k

**Case 2**

If t is negative, then t will appear in an odd position in the sequence.

Thus,

f(2k-1)=-\frac{(2k-1)-1}{2}

=t

With

t=-k

This implies that for every negative value t in \mathbb{Z} there is a natural number 2k-1

Therefore, f is surjective

Hence, \mathbb{Z} is countable set.

Now we define the sequence

a_n=n-100

>-100

For n \in \mathbb{N}

This is clearly a subset of \mathbb{Z} and so is countable.