Let $A$ be a nonempty finite subset of $\mathbb{R}$.
Firstly, let me write down how to define the term "the smallest element of $A$" formally.
Suppose 'for every $x\in A$, there exists $y \in A$ such that $x>y$'.
Since $A$ is nonempty, we can pick an element $x_0\in A$. Then by assumption, there exists $x_1$ such that $x_0>x_1$. Continuing in this manner, we get a finite sequence $x_0>x_1>...>x_{|A|}$. This means that $\{x_0,...,x_{|A| }\}$ is a subset of $A$, but the former set has cardinality greater than the latter set. This is a contradiction.
Thus, there exists $x\in A$ such that $x\leq y$ for all $y\in A$.
And uniqueness of such $x$ follows easily. Hence, we just defined the term "the smallest element of the set $A$" formally. Note that this is the minimum among distict elements. If we continue in this manner, we can define the $j$-th minimum among distinct elements formally. (For example, $2$-nd minimum can be defined as $min ( A\setminus\{min A\})$)
Now my question is the below:
Say, we have defined the term "$j$-the smallest element" (possibly not disticint) although we have not defined it formally yet because all we know what it means informally. Denote this function by $F_j:\mathbb{R}^n \rightarrow \mathbb{R}$.
This is slightly different situation from previously described one, because the case $x_1=...=x_n$ may happen. However for the case $j=1$, there is actually no difference from the above distinct-case. However, for $j>1$, I am not sure how to define it formally. How do we define it?
One way is the following:
Let $x\in \mathbb{R}^n$
Then, there exists permutation $\sigma\in S_n$ such that $x_{\sigma_1}\leq\cdots\leq x_{\sigma_n}$.
Let $\tau \in S_n$ be another permutation with the same property.
If one can show that $x_{\sigma_i}=x_{\tau_i}$ for all $i$, then $F_j$ is well-defined and we are done. How do I prove this equality?
And is there another clever "neat" way to define $F_j$ formally?
Thank you in advance!
[EDIT]
My description seems very confusing. To make it clear, take, for example, $a_1=1, a_2=1, a_3=2$. I want to have $1$ as both the first and second smallest element and $2$ as the third smallest element, so that $F_1(1,1,2)=1, F_2(1,1,2)=1, F_3(1,1,2)=2$.
A formulation that works for your other question also.
Let $S$ be the collection of the $\binom nj$ subsets of $\{1,2,\ldots,n\}$ that have exactly $j$ elements.
To each $M\in S$ define $$ \max M=\max\{x_i\mid i\in M\}. $$ Then the $j$th smallest number $z_j$ is $$ z_j=\min\{\max M\mid M\in S\}. $$
After all, if $z_1\le z_2\le \cdots\le z_j\le\cdots$ are the numbers $x_i$ rewritten in ascending order, then the choice of $M$ corresponding to $\{z_1,z_2,\ldots,z_j\}$ has $z_j$ as $\max M$. And obviously that is (one of) the smallest you can get, i.e. $\min\{\max M\}$.