I have the following question :
Defintion : $R \subseteq \mathbb{N}^\mathbb{N} \times \mathbb{N}^\mathbb{N}$ two functions $f_1,f_2 \in \mathbb{N}^\mathbb{N}$ are almost identical if $A \subseteq \mathbb{N}$ (A is infinite).
doesn't exist such that for all $i\in A$ applies $f_1(i)\neq f_2(i)$
if $f_1,f_2 \in $ are almost identical then $(f_1,f_2)\in R$
- Proof $h\in \mathbb{N}^\mathbb{N}$ proof $[h]_r$ is infinite countable.
What I did
I try to show an injective f:$\mathbb{N} \rightarrow [h]_r$ and also an injective $g:[h]_r \rightarrow \mathbb{N}$ and from cantor bernstein theorem there's a function that is injective and subjective so we know $[h]_r$ is infinite countable.
I thought of function $f$ as following since $(i\in \mathbb{N})$ $[h]_r=\{{j_i\in \mathbb{N}^\mathbb{N} |(h,j_i)\in R\}}$
$$f(i)=j_i (i\in \mathbb{N})$$
injective :
let $i,k \in \mathbb{N}$ such that $k\neq j$ then $f(i)=j_k\neq j_j=f(j)$ since there's $n\in \mathbb{N}$ that $j_k(n)\neq j_i(n)$.
$g:[h]_r\rightarrow \mathbb{N}$
for all $a,b \in [h]_r$ exists $i\in \mathbb{N}$ such that $a(i)\neq b(i)$ let $m=\max\{{i | a(i)\neq b(i) \}}$
for all $a\in [h]_r$ $$f(a)=p_1^a(1)*...*p_m^a(m)$$ while $p_1=2,p_2=3,..$ (prime numbers)
injective
from the prime number theorem each number has a unique way to be written.
$g,f$ are injective so from cantor bernstein theorem we know that $[h]_r$ is infinite countable.
I'm not sure about if i defined function $f$ correctly, I'd like to have an option about it.
Thanks in advence!
$\newcommand{\NN}{\mathbb{N}}$ $\newcommand{\ZZ}{\mathbb{Z}}$I'm having some trouble reading your arguments. Please don't use symbols that you are yet to define and clearly indicate quantifiers whenever you use variables.
Your approach is correct though. We show $|\NN|\le|[h]|$ and $|[h]|\le|\NN|$.
The first inequality is simple enough. Define $f:\NN\to[h]$ as follows. For each $k\in\NN$, define $f(k):\NN\to\NN$ by $$ f(k)(n) = \begin{cases} h(n)+1 & \text{if}\ n=k, \\ h(n) & \text{if}\ n\ne k. \end{cases} $$ This mapping is well-defined because for each $k\in\NN$ the function $f(k)$ is almost identical to $h$ (i.e., $f(k)\in[h]$ for every $k\in\NN$). The function $f$ is an injection because $k\ne j$ implies $$f(k)(j)=h(j)\ne h(j)+1=f(j)(j),$$ which shows that the functions $f(k)$ and $f(j)$ are distinct.
It remains to prove $|[h]|\le|\NN|$. It's not immediately clear to me how one would define an injection from $[h]$ into $\NN$. However, note that every function in $[h]$ must agree with $h$ at all but finitely many points. Thus to each element $a\in[h]$ we can find a finite subset $N_0$ of $\NN$ and a function $z:N_0\to\ZZ$ such that $$ a(n) = \begin{cases} h(n)+z(n) & \text{if}\ n\in N_0, \\ h(n) & \text{if}\ n\not\in N_0. \end{cases} $$ There are only countably many finite subsets of $\NN$ and, for each finite subset $N_0$ of $\NN$, there are only countably many functions from $N_0$ into $\ZZ$. This shows that $[h]$ is countable.
To write this argument formally, we let $F(\NN)$ denote the set of finite subsets of $\NN$ and define a function $$g:[h]\to\bigcup_{N_0\in F(\NN)}\ZZ^{N_0}$$ as follows. Given $a\in [h]$, there exists a finite set $N_0$ of natural numbers such that $a(n)\ne h(n)$ iff $n\in N_0$. We define $g(a):N_0\to\ZZ$ by $g(a)(n)=a(n)-h(n)$. Then $g(a)\in\ZZ^{N_0}$ so that $g$ is well-defined.
To see that $g$ is one-to-one, suppose $a\ne b$ are in $[h]$. We have two cases. If there exists $n\in\NN$ such that $a(n)\ne h(n)$ and $b(n)=h(n)$ (or similarly $a(n)=h(n)$ and $b(n)\ne h(n)$), then the functions $g(a)$ and $g(b)$ are defined on different sets and thus $g(a)\ne g(b)$. Otherwise, suppose $a$ and $b$ differ from $h$ at the same points. Since $a\ne b$, there exists $n\in\NN$ such that $a(n)\ne b(n)$. Consequently $g(a)(n)=a(n)-h(n)\ne b(n)-h(n)=g(b)(n)$, which proves $g(a)\ne g(b)$. Therefore $g$ is an injection.
This shows that $$ |[h]| \le \left|\bigcup_{N_0\in F(\NN)}\ZZ^{N_0}\right|. $$ Since $\cup_{N_0\in F(\NN)}\ZZ^{N_0}$ is a countable union of countable sets, we conclude that $|[h]| \le |\NN|$.