Finite set with $\sup\{S\} \notin S $

Question

Just a quick question about sets.

Can anyone here think of an example of a finite set $S$ where the $\sup\{S\} \notin S $ ?

I found this on a past exam paper for college, and am unsure if there exists a finite set with this property.

Answer 1

If you're talking of a subset of real numbers then it can't be: if $\;S=\{s_1,...,s_n\}\;$ then we can actually order the elements of $\;S\;$ (all the time wrt the usual order), say $\;S=\{s_{i_1},...,s_{i_n}\}\;$ , with $\;s_{i_1}<s_{i_2}<...<s_{i_n}\;$ , and then clearly $\;\sup(S)=s_{i_n}\in S\;$

Answer 2

The answer depends on the context. Is the following example allowed?

Consider a set $X$, and its power set $\mathcal{P}(X)$, ordered by inclusion. It admits a supremum, where $\sup\{A, B \} = A \cup B$, for $A, B \subseteq X$.

Now if you take $X = \{1, 2, 3\}$, $A = \{1,2\}$, $B = \{1, 3\}$ and $S = \{A, B\}$, then $\sup(S) = X \notin S$.

Answer 3

Every finite linear order admits a maximum, so if $S$ is linearly ordered, $\sup S=\max S\in S$.

Therefore the ambient order cannot be a linearly ordered set, and $S$ itself must witness that. Knowing this we can now have plenty of examples:

1. $S=\mathcal P(X)\setminus\{X\}$ where $X$ is a finite set with at least two elements, and the order is $\subseteq$.

2. Order the natural numbers by divisibility, $k\preceq m\iff k\mid m$, and take $S$ to be any finite set of prime numbers with at least two elements.

3. Consider the order on $\Bbb N\times\{0,1\}$ defined by $(m,i)\preceq(n,j)\iff m\leq n\land i\leq j$. Then take the set $\{(n+1,0),(n,1)\}$.