The set of integers is countable

[evinda]

Hello! (Smirk)

Proposition
The set $\mathbb{Z}$ of integers is countable.

Proof

$\mathbb{Z}$ is an infinite set since $\{ +n: n \in \omega \} \subset \mathbb{Z}$.

$$+n= [\langle n, 0 \rangle]=\{ \langle k,l \rangle: k+n=l\}$$

We define the function $f: \omega^2 \to \mathbb{Z}$ with $f(\langle n,m \rangle)=m-n$ that is obviously surjective.
Since $\omega^2$ is countable we conclude that $\mathbb{Z}$ is countable.In this case can we subtract two natural numbers since we define $+n= [\langle n, 0 \rangle]=\{ \langle k,l \rangle: k+n=l\}$ ?Also how could we show that $f$ is surjective?
It suffices to show that $\forall y \in m-n, \exists \langle m, n \rangle $ such that $f(\langle m, n \rangle)=m-n$, right?
How could this be shown? (Thinking)

 

[Euge]

Hi evinda,

What do you mean when you say that $\Bbb Z$ is a finite set? As for the surjectivity of $f$, take any integer $l\in \Bbb Z$. If $l < 0$, then $f(1,1 - l) = l$. If $l = 0$, then $f(1,1) = l$. Finally, if $l > 0$, then $f(l+1,1) = l$.

 

[evinda]

What do you mean when you say that $\Bbb Z$ is a finite set?

That was a typo... (Tmi)

As for the surjectivity of $f$, take any integer $l\in \Bbb Z$. If $l < 0$, then $f(1,1 - l) = l$. If $l = 0$, then $f(1,1) = l$. Finally, if $l > 0$, then $f(l+1,1) = l$.

How did you calculate these values? (Thinking)

 

[Euge]

I might be misunderstanding your question, but I calculated the values using the formula for $f$.

 

[evinda]

I might be misunderstanding your question, but I calculated the values using the formula for $f$.

You pick any $l \in \mathbb{Z}$. If $l<0$ why does it hold that $m=1, n=1-l$?

 

[Euge]

Well, if $l < 0$, then $1-l$ is a natural number, and $ f (1,1-l) = 1-(1-l) = l $.