$n^{th}$ minimum/maximum notation

Question

Given a set $S$ of comparable items, such that $|S|=m$, can I say that the $n^{th}$ minimum (maximun), say $a$, is:

$$a=\min_m \{S\}$$

where $n \leqslant m$.

If not, which is the correct notation?

Answer 1

I think the descriptive way is the best. You can define $\min\limits_n S$ as "the $n$-th smallest element of $S$" if you want, but you still need that description.

However, it might be less confusing to define $m(S,n)$ as "the $n$-th smallest element of $S$", since $\min\limits_n S$ is a standard notation for "minimum of $S$ over all possible values of $n$".

If you want to mess things up by avoiding the description completely, then you can define

$$f:\ \mathbb{N} \to S, \quad f(k) = \begin{cases} \min S, & k = 1, \\ \min \{s:\ s \in S, s > f(k-1)\}, & k > 1. \end{cases}$$

Of course, I advise against it, because it overly cryptic and the descriptive way works just fine.

A (sometimes wrong) alternative

You might consider saying

Let $S = \{ s_1, s_2, \dots \}$, where $s_i < s_j$ for all $i < j$.

and then dealing with $s_n$, which is obviously the $n$-th smallest value of $S$. But, this is generally wrong, as it assumes that $S$ is enumerable. As an example, consider $S = \mathbb{N} \cup [ x, y ]$ for some $x < y$. Here, the first $n = \lfloor x \rfloor$ minimal numbers are well defined, but you cannot write $S$ in the above manner.

If $S$ is finite or infinite but enumerable, the above notation may be useful (and is used fairly often, for example for $n$-th smallest/biggest singular value of a given matrix).

However, a side description is still useful. For example, if using the above notation, I'd still write something like

... Let $s_n$ be the $n$-th smallest element of $S$. ...

That way, you point the reader in the right direction, while also being mathematically precise.