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.
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2See [Every finite set...](http://math.stackexchange.com/questions/259893/everyfinitesetcontainsitssupremumproofimprovement) – J. W. Perry Nov 02 '13 at 08:49

2Suppose there is no maximal element then what happens? For every element $x \in S$,you can always find $ y \in S$ such that $x
– wannadeleteacct Nov 02 '13 at 08:49 
3You mean a **nonempty** finite set? – bof Nov 02 '13 at 09:11

isn't this weierstrass theorem? – Ant Nov 02 '13 at 13:31

See also http://math.stackexchange.com/questions/24996/questionrelatedwithpartialorderfinitesetminimalelement – Martin Sleziak Feb 13 '15 at 18:32
5 Answers
Let $S = \{s_1, \ldots,s_n\}$ be a nonempty finite set of size $n > 0$. 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 $s_{k+1} < m'$. 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: $$ m = s_{k+1} < m' \leq s' \leq M' =M $$
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 $s_{k+1} > M'$. 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: $$ m = m' \leq s' \leq M' < s_{k+1} = M $$
Hence, we have shown that $S$ has a minimum and maximum element, as desired.
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I see where you are coming from, and your proof is through. Thank you for taking the time for answering this. However, I am not really sure whether I can use proof for induction when such elements are not in $\mathbf{N}$. Some time in the past, I tried to use a similar method for proving a statement about $\mathbf{R}$ and was told that I couldn't use proof by induction. I found this link which says otherwise and is pretty through: http://math.stackexchange.com/questions/4202/inductiononrealnumbers. Would you mind elaborating on why do you think induction would work in this case? – CoffeeIsLife Nov 02 '13 at 21:09

5It is true that you cannot use induction on a set where wellordering does not hold, such as the real numbers. However, notice that my proof uses induction on $n \in \mathbb N$, and induction holds for the natural numbers due to wellordering. We are NOT inducting on the elements of $S$ (for all we know, each $s_i \in S$ could be real numbers, or matrices, or purple hippos). – Adriano Nov 03 '13 at 01:48

Are you then implying that a finite set on $\mathbf{R}$ does not always have a maximum and a minimum? – CoffeeIsLife Nov 03 '13 at 21:13

7No, I'm not. As I have proven, finite sets on $\mathbb R$ always have a maximum and a minimum. I suspect that you are confusing the set that we are induction on (that is, the natural numbers, since finite sets always have a cardinality that belongs to $\mathbb N$) with the set that the elements of $S$ belong to (say, the real numbers). – Adriano Nov 04 '13 at 11:52

I have a question here: Why the finitude of the set forces the existence of the minimal and maximal elements? – Rosa Maria Gtz. Feb 04 '17 at 19:05

A set S being finite means that there is a bijection between its elements and a natural number. If every pair of elements of S can be compared, then they can be ordered by the elements of a natural number. Every n+1= {0,1,..., n}, so ordering S, just says there's a 0th and an nth element which is exactly what we want. He assumes that S'=S{x} has a maximal and a minimal element. He then compares x with, the maximum and the minimum, it either is a new maximum or a new minimum, or it itsn't, but S always has a maximum and minimum. This works because of induction regardles of the nature of S. – Ricardo Acuna Dec 06 '18 at 20:12

1I guess the proof here assumes there's an order relation on the elements of S. – Ricardo Acuna Dec 06 '18 at 20:14
Let $F$ be a finite set. if $F$ is $\{x\} $ then we are done since we vacouly have $x \geq x $ and hence $x = \max \{ x \} $. If $F = \{ a_1 ,... a_n \} $. assume they are different, otherwise we are back again to singleton case. Now, take $a_1$. IF $a_1 $ is greater than any other $a_i$ then set $a_1 = \max F $ and we are done. IF not, take $a_2$, and repeat previous step. Continue in this manner inductively. Eventually, we get the max.
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I would just say based on Suppes, that use of induction to prove this conjecture begs the question for the following reasons:
 The minimality or maximality of the elements within a set, A, cannot be defined without specifying an ordering relation, R, on A.
 Different R's will yield different minimal elements, called Rminimal elements.
 Set A only has a unique Rminimal element iff all its nonempty subsets have unique Rminimal elements, R is connected, and R is asymmetric. (i.e. R is a well ordering)
 Having a unique Rminimal element does not guarantee a unique Rmaximal element unless A is finite.
Set A is finite iff every nonempty family of subsets of A has a minimal element [ordered by strict inclusion '$\subset$'] A. Tarski via Suppes
Furthermore, if one such family of subsets, F, has a minimal element , then it must also have a maximal element via the following argument:
If F has a minimal element, then construct G from the elements of F where z is an element of G iff for some y in F, z is Union F  y. If z is minimal in G, then y is maximal in F, and vice versa. If F does not have a maximal element, then G cannot have a minimal element, but G does have a minimal element, hence contradiction, and the theorem is proved.
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Sorry, but this proves nothing. Besides, proving the statement in the question by induction ***is*** the way to go. – egreg Feb 01 '19 at 12:59

Sorry, but induction is actually proved from this theorem, so using it begs the question. – xxxx0xxxx Feb 01 '19 at 20:06

Suppes uses Tarski's definition of a finite set, which is: A set is finite iff all its nonempty families of subsets have a minimal element. Thus G and F must have minimal elements if the set is finite. But if one doesn't have a maximal element, then the other can't have a minimal element, which contradicts Tarski's definition. – xxxx0xxxx Feb 01 '19 at 20:22

And I quote from Suppes et al. "The proof of the theorem about induction for finite sets is facilitated by having available the already defined notion of maximal element of a family of subsets." – xxxx0xxxx Feb 01 '19 at 20:49

There's no hint in the question that Tarski's definition is used and I find it unlikely. Anyway, the argument you sketch should be better written. – egreg Feb 01 '19 at 21:24

Then what makes a set finite? Tarski's definition is the only one that does not require choice, ordinals, or induction. Minimality and maximality define finite. – xxxx0xxxx Feb 01 '19 at 22:13
One needs to be careful about the set where to work, and the definition of the ordering. If the set is well ordered, then the above proof works fine. Otherwise, it can happen that there is a supremum, but not a maximum. We can, for example, consider the partial ordering on the set of $2\times 2$ realvalued matrices, where two matrices $A$ and $B$ are such that $A\le B$ if and only if all the entries of $BA$ are nonnegative.
With the abovementioned order relation and set, we consider the subset $$ E = \left\{ \begin{pmatrix} 1 & 0 \\ 0& 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0& 1 \end{pmatrix} \right\} .$$
We can see that $E$ is bounded from above (by $ \begin{pmatrix} 1 & 0 \\ 0& 1 \end{pmatrix}$ for example), and from below( by $\begin{pmatrix} 0 & 0 \\ 0& 0 \end{pmatrix}$ for example). However, the two elements of $E$ are not comparable using the defined order relation, so none of them is the maximum.

If you really want to argue this silly point, a simpler and clearer example is $\{ \text{potato}, \text{bicycle}\}$. – MJD Sep 03 '21 at 13:14
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.
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1How do you know that $m$ and $M$ even exist? This *exactly* what the question asks. – Asaf Karagila Nov 23 '15 at 20:48

This is standard stuff, the min and max functions for a finite number of arguments is well defined – Georgy Nov 30 '15 at 13:48

1How do you prove they are welldefined? The question is *essentially* requiring to show that you can do that. The answer by ILoveMath use an approach similar to yours, but with an actual proof that this is welldefined for any finite number of arguments. You can't just reduce something to an equivalent problem and call it a proof, especially if you don't actually solve that equivalent problem. – Asaf Karagila Nov 30 '15 at 13:49

ok I define them:$$ \min(s_1,s_2,\dots,s_n)=\min(\min(s_1,\dots,s_{n1}),s_n) $$ similarly for the maximum. – Georgy Jul 30 '16 at 17:59

1Yes, that's fine, but it *is* an inductive definition, and the question is about a rigorous proof that a finite set has minimum and maximum. So the inductive argument needs to be brought up. You can't avoid it, and your answer sweeps this under the rug. – Asaf Karagila Jul 30 '16 at 18:02


2Which appears in details in the two answers there were there two years before you posted your answer; and does not appear in your answer at all. – Asaf Karagila Jul 30 '16 at 18:06

My point is that in every proof there are some things that we consider trivial and these are the starting point (see Richard Feynman). In my opinion the min and max functions don't need any introduction. Indeed one might question if we can define them for any finite set. What's the min({1},{1,2})? You didn't specify "set of real numbers". But even if you do that one can ask how are real numbers defined and so on. Some things are just assumed and I assumed that. I would allocate full marks in an exam if the student mentioned that. – Georgy Jul 30 '16 at 18:22

3Sure, and I wouldn't expect anyone to prove to me that this is welldefined when I referee a paper. I would, however, expect someone who is asking *why* a finite set has a minimum and maximum (and clearly, the real analysis tag means this is a set of real numbers) expect the answer to be more than "well, just apply the minimum and maximum functions to the set" which is circular and terrible. And if someone would have written me this answer in a set theory exam, they would have received exactly 0 points for it. – Asaf Karagila Jul 30 '16 at 19:31