I have proven that if $X $ is a finite set and $Y$ is a proper set of $X$, then does not exist $f:X \rightarrow Y $ such that $f$ is a bijection.
I'm pretending to show that the naturals is an infinite set. Let $P=\{2,4,6…\} $. So, by contraposity i just have to show that $f: \mathbb{N} \rightarrow P$, given by $f(n)=2n$, is actually a bijection.
Logically speaking (my book), i have yet constructed only the natural numbers, its addition and multiplication.
By far, i have shown that $f$ is injective. But in order to prove that $f$ is a surjection, how can i do that without using the usual: "Take any $y\in P$. Given $x=\frac{y}{2}$ …"? Because $x$ is actually a rational number given in a "strange form", which I yet didn't construct.
Best Answer
You are not using the definition of even number.
Definition 1 (Even number). We say that a natural number $n$ is an even number if $n=2k$ for some natural number $k$.
In order to prove that $f\colon P\to\mathbb N$ is a surjection, let $y\in P$. Now we need to find some $x\in\mathbb N$ such that $f(x)=2x=y$. But such $x$ exists by Definition 1 as desired.