Proof of injective functions

Question

Let $f:\mathbb{N}\to\mathbb{N}$ and $N_k=\{1,2,3,...,k\}$ for all $k\in\mathbb{N}$.

I have to prove two statements. If $f_{\big|N_k}$ is injective function for all $k\in\mathbb{N}$, then $f$ is injective function(one to one) and second if $f[N_k]=N_k$ for all $k\in\mathbb{N}$, then $f$ is identity function.

Note that $f_{\big|N_k}$ is restricted domain of function and $f[N_k]=N_k$ is image of function.

I prove the first statement, let $x_1,x_2\in\mathbb{N}$ and assume that $f(x_1)=f(x_2)$. If $x_1,x_2\in{N_k}$, then we are done but if not ? Next, the second statement I have to show that $f(x)=x$ for all $x\in\mathbb{N}$ but how to show that ?

Tanks for someone to help me.

Answer 1

If $x_1,x_2\in\Bbb N$, take some natural number $k$ such that $x_1,x_2\leqslant k$. Then $x_1,x_2\in N_k$ and so, since $f|_{N_k}$ is injective, if $f(x_1)=f(x_2)$, then $x_1=x_2$.

Concerning the other question, suppose the $f$ is not the identity function. Then there is a smallest natural $k$ such that $f(k)\ne k$. The number $k$ cannot be $1$, since $f(N_1)=N_1$, which means that $f(\{1\})=\{1\}$, or $f(1)=1$. So, $k>1$. But $f(N_{k-1})=N_{k-1}$ and $f(k)\ne k$ implies that $f(N_k)\ne N_k$.

Answer 2

1. Consider two natural numbers not equal to each other. Label the smaller one $j$ and the larger one $k$. $j< k$ so $j, k \in \mathbb N_k$. $f|_{\mathbb N_k}$ is injective so $f(j) \ne f(k)$. But $j$ and $k$ were arbitrary so for any $j,k \in \mathbb N$ if $j\ne k$ then $f(j) \ne f(k)$ and $f$ is injective.

2. Let $n \in \mathbb N$. $n\in \mathbb N_n$ so $f(n) \in \mathbb N_n$ so $f(n) \le n$.

If $f(n) < n$ then let $f(n) = k < n$. But $f(\mathbb N_k) = \mathbb N_k$ so there is an $m \in \mathbb N_k$ so that $f(m) = k = f(n)$. But by 1) $f$ is injective so $m = n$. But $n \not \in \mathbb N_k$. So $f(n) \not < n$ and $f(n)\le n$ and so $f(n) = n$.