# Proving that the set of the natural numbers is infinite

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.

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.

Definition 2 (Surjective functions). A function $f$ is surjective if every element in $Y$ comes from applying $f$ to some element in $X$: $$\text{For every }y\in Y\text{, there exists }x\in X\text{ such that }f(x)=y.$$

The function $f$ you gave is surjective onto its image, and this image is a proper subset (can $2n=1$ be realized by some $n \in\mathbb{N}$?).

By the fact that there is a bijection $f: \mathbb{N} \to 2\mathbb{N}$, and $2\mathbb{N}$ is a proper subset of $\mathbb{N}$, it follows that $\mathbb{N}$ cannot be finite.

Showing there is a bijection $f: \mathbb{N} \to \mathbb{N}$ doesn't do much, because such a bijection does not contradict the theorem you've mentioned.

• Yes, i've edited my post. But how can i actually show that $f$ from the naturals to the even naturals is a bijection? – math.h Jan 9 '17 at 0:18
• If $m$ is even, then $m=2n= f(n)$. This shows surjectivity, and you've already shown injectivity. – Hayden Jan 9 '17 at 0:35
• Thank you. Now it is really clear for me. – math.h Jan 9 '17 at 0:37