A finite set always has a maximum and a minimum. I am pretty confident that this statement is true. However, I am not sure how to prove it. Any hints/ideas/answers would be appreciated.
Let $S = \{s_1, \ldots,s_n\}$ be a nonempty finite set of size $n &gt; 0$. We will show by induction on $n \in \mathbb N$ that there exist some $m,M \in S$ such that for all $s \in S$, we have that $m \leq s \leq M$.

Base Case: For $n=1$, we have $S = \{s_1\}$, so taking $m = s_1$ and $M=s_1$ trivially satisfies the required condition.

Induction Hypothesis: Assume that the claim holds for $n=k$, where $k \geq 1$.

It remains to prove that the claim holds true for $n = k+1$. To this end, choose any set $S$ with $k+1$ elements, say $S = \{s_1 ,\ldots,s_k,s_{k+1}\}$. Now by the induction hypothesis, the subset:
$$
S' = S \setminus \{s_{k+1}\} = \{s_1 ,\ldots,s_k\}
$$
has a minimum element and a maximum element. That is, we know that there exists some $m',M' \in S'$ such that for all $s' \in S'$, we have that $m' \leq s' \leq M'$. Now observe that $s_{k+1}$ must fall under $1$ of $3$ cases:

Case 1: Suppose that $s_{k+1} &lt; m'$. Then take $m = s_{k+1}$ and $M=M'$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and:
$$
m = s_{k+1} &lt; m' \leq s' \leq M' =M
$$

Case 2: Suppose that $m' \leq s_{k+1} \leq M'$. Then take $m = m'$ and $M=M'$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and:
$$
m = m' \leq s_{k+1} \leq M' = M
$$
$$
m = m' \leq s' \leq M' = M
$$

Case 3: Suppose that $s_{k+1} &gt; M'$. Then take $m =m'$ and $M=s_{k+1}$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and:
$$
m = m' \leq s' \leq M' &lt; s_{k+1} = M
$$

Hence, we have shown that $S$ has a minimum and maximum element, as desired.
I suppose you mean a set of numbers, $S=\{s_1,s_2,\dots,s_n\}$ and $n$ is the finite size, just let

$$
m=\min(s_1,s_2,\dots,s_n)
$$
$$
M=\max(s_1,s_2,\dots,s_n)
$$
Q.E.D.