How to define an injective function from a set of sets of length $n$ to a set of sequences of length $n$ ($\mathbb N\times\mathbb N\times...\times \mathbb N$)?
As far as i understand a set of sets of length $n$ is something like this (for example if $n=3$): $$ \{\{1,2,3\}, \{5,7,9\}, \{100,345,500\},...,\} $$ while a sequence of $\mathbb N\times \mathbb N\times \mathbb N $ is something like this: $$ (1,2,3), (5,7,9), (100, 345, 500),... $$ such that the main difference between the sets and sequences is that order matters in sequences while it doesn't in sets.
So I could say that if $S_n$ is a set of sets of length $n$ then let: $$ f:S_n\to \mathbb N\times\mathbb N\times...\times \mathbb N $$ such that we order the elements in each set $S_i, 1\le i\le n$ in increasing order. Intuitively I understand that $f$ is injective but how one really does prove it?
Injectivity is best shown by exhibiting aright inverse. First lets formalize your $f$. Let $$\mathcal A=\{\,A\subseteq \Bbb N\mid |A|=n\,\}$$ be the set of length $n$ sets of natural numbers. Let $$\mathcal B=\{\,b\mid b\text{ is a map }\{1,\ldots,n\}\to\Bbb N\,\}$$ be the set of length $n$ sequences of natural numbers. For convenience, we consider the subset of injective maps $$ \mathcal B_0:=\{\,b\in\mathcal B\mid b\text{ is injective}\,\}.$$
We can define $f\colon \mathcal A\to\mathcal B$ as follows: Given $A\in\mathcal A$, let $f(A)$ be given by $f(A)(k)=$ the $k$th smallest element of $A$, for $k\in\{1,\ldots,n\}$.Or recursively, $f(A)(k):=\min (A\setminus \{\,f(A)(j)\mid 1\le j<k\,\})$. One still has to show that this really defines a map, but I think there is a simpler way to begin with:
An alternative (and in this context more useful) way of describing $f$ is this: $\mathcal B$ is linearly ordered by lexical order: If $b_1,b_2\in\mathcal B$ with $b_1\ne b_2$, let $k=\min\{\,j\in\{1,\ldots,n\}\mid b_1(j)\ne b_2(j)\,\}$ and we say $b_1<b_2$ if $b_1(k)<b_2(k)$. In fact, one verifies that this is a well-order on $\mathcal B$ (essentially because $<$ is a well-order on $\Bbb N$). Now given $A\in\mathcal A$, the fact that $|A|=n$ tells us that there exists at least one bijection $\{1,\ldots,n\}\to A$, which can be viewed as an element $b\in \mathcal B$, in fact it is clear that $b\in\mathcal B_0$. This allows us to define $f\colon \mathcal A\to\mathcal B$ (or in fact even $f\colon \mathcal A\to\mathcal B_0$) by setting $$ f(A):=\min\{\,b\in\mathcal B_0\mid \operatorname{im}(b)=A\,\}.$$ Now in order to show that $f\colon \mathcal A\to\mathcal B_0$ is injective, we can simply exhibit a map $\mathcal B_0\to \mathcal A$ such that $g\circ f=\operatorname{id}_{\mathcal A}$. Quite obviously. the straightforward map given by $$ g(b):=\operatorname{im}(b)$$ does the trick.