Question

Show that the set of odd integers is countable

Short answer

Hence, set of odd integers is countable.

Step-by-step solution

Countable set

A set is countable if it is finite or countably finite

Set of odd integers is countable.

Given:S is the set of odd integers.

To proof: S is countable.

Proof: S is not finite, thus we need to proof that S is countable infinite.

Let us definite the function f as:

$f:S\to Z^{+},f\left(n\right)=\left\{\begin{array}{l}n\\ -n+1\end{array}\right.$

f Is one-to-one: Let$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{b}\right)$

If a and b are positive, then by definition of

If a and b are negative, then by definition of$\mathrm{f}:-\mathrm{a}+1=-\mathrm{b}+1$ which implies a=b

If a positive and b negative, then by definition of $\mathrm{f}:\mathrm{a}=-\mathrm{b}+1$, which is impossible for odd integers a and b.

If a negative and b positive, then by definition of $\mathrm{f}:\mathrm{b}=-\mathrm{a}+1$, which is impossible for odd integers a and b.

Thus$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{b}\right)$ implies a and b both positive or both negative and then also implies a=b, which means that is one-to-one.

One to one correspondence

IfIs onto: Let$\mathrm{b}\in {\mathrm{Z}}^{+}$

If b odd, then $\mathrm{a}=\mathrm{b}\in \mathrm{S}$with$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{b}\right)=\mathrm{b}$

If b even, then role="math" localid="1668423944271" $\mathrm{a}=-\mathrm{b}+1\in \mathrm{S}$withrole="math" localid="1668423955026" $\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}(-\mathrm{b}+1)=-(-\mathrm{b}+1)=\mathrm{b}-1+1=\mathrm{b}$

Thus, f is onto.

Since, f is onto and one-to-one, f is a one-to-one correspondence from S to the positive integers. This means that S is infinitely countable and thus S is countable.