Prove that the set of all even integers is denumerable

[issacnewton]

Homework Statement

Prove that set of all even integers is denumerable

Relevant Equations

A set $A$ is called denumerable if $\mathbb{Z}^+ \sim A$

Now, set of even integers is $A = \{ \cdots, -4, -2, 0, 2, 4, \cdots \}$. We need to prove that $\mathbb{Z}^+ \thicksim A$. Which means that, we need to come up with a bijection from $\mathbb{Z}^+$ to $A$. We know that $\mathbb{Z}^+ = \{1,2,3,\cdots \}$. I define the function $f : \mathbb{Z}^+ \longrightarrow A$ as follows.
$$ f(x) =
\begin{cases}
x-1 & \text{if } x \text{ odd} \\
-x & \text{if } x \text{ even}
\end{cases}
$$
Now, we will prove this is both one-to-one and onto. Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in \mathbb{Z}^+$. From the way we have defined this function, we either have $f(x_1) \geqslant0$ or $f(x_1) < 0$. Consider that $f(x_1) \geqslant0$. It also follows that $f(x_2) \geqslant 0$. And we have $f(x_1) = x_1 - 1 = f(x_2) = x_2 - 1$. And it follows from this that $x_1 = x_2$. Other case is that $f(x_1) < 0$. From this, we have $f(x_2) < 0$. And using the function definition, we have, $f(x_1) = -x_1 = f(x_2) = - x_2$. It follows that $x_1 = x_2$. So, it has been proved in both cases that $x_1 = x_2$ and since $x_1, x_2$ are arbitrary to begin with, it follows that the function is one-to-one. Is this reasoning ok so far ?

$\spadesuit\spadesuit\spadesuit\spadesuit$

 

[PeroK]

Homework Statement:: Prove that set of all even integers is denumerable
Homework Equations:: A set $A$ is called denumerable if $\mathbb{Z}^+ \sim A$

Now, set of even integers is $A = \{ \cdots, -4, -2, 0, 2, 4, \cdots \}$. We need to prove that $\mathbb{Z}^+ \thicksim A$. Which means that, we need to come up with a bijection from $\mathbb{Z}^+$ to $A$. We know that $\mathbb{Z}^+ = \{1,2,3,\cdots \}$. I define the function $f : \mathbb{Z}^+ \longrightarrow A$ as follows.
$$ f(x) =
\begin{cases}
x-1 & \text{if } x \text{ odd} \\
-x & \text{if } x \text{ even}
\end{cases}
$$
Now, we will prove this is both one-to-one and onto. Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in \mathbb{Z}^+$. From the way we have defined this function, we either have $f(x_1) \geqslant0$ or $f(x_1) < 0$. Consider that $f(x_1) \geqslant0$. It also follows that $f(x_2) \geqslant 0$. And we have $f(x_1) = x_1 - 1 = f(x_2) = x_2 - 1$. And it follows from this that $x_1 = x_2$. Other case is that $f(x_1) < 0$. From this, we have $f(x_2) < 0$. And using the function definition, we have, $f(x_1) = -x_1 = f(x_2) = - x_2$. It follows that $x_1 = x_2$. So, it has been proved in both cases that $x_1 = x_2$ and since $x_1, x_2$ are arbitrary to begin with, it follows that the function is one-to-one. Is this reasoning ok so far ?

$\spadesuit\spadesuit\spadesuit\spadesuit$

Yes. Looks fine.

 

[issacnewton]

Thanks PeroK