Finding the minimum and the maximum when speaking of relations I'm trying to figure out how to find the minimum and the maximum when speaking of relations. Consider $A=(2,3,\ldots,13)$. Also consider the following relation $R$:
$$
(x,y)\in R \Leftrightarrow x|y
$$
where $x,y\in A$. I need to find the minimum and the maximum of the relation. What does it mean? How do I find them?
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\newcommand{\p}{\preccurlyeq}
$
A &quot;maximal&quot; element of a set $A$ with partial order $\p$ is an $M \in A$ such that there is no $a \in A$ where both $M \p a$ and $M \ne a$. Likewise, a minimal element of $(A,\p)$ is an $m \in a$ such that no $a \in A$ satisfies $a \p m$ and $a \ne m$ together.
Consequently, $a \in A$ is not maximal if there exists $b \in A$, $b\ne a$, such that $a \p b$. Similarly, $a$ is not minimal if $b \p a$ but $b \ne a$.
So, run through the list invalidating candidates for maxima first:

$\not \exists x \in A$ such that $x \mid 2$ (except itself)
$\not \exists x \in A$ such that $x \mid 3$ (except itself)
$2 \mid 4$
$\not \exists x \in A$ such that $x \mid 5$ (except itself)
$3 \mid 6$
$\not \exists x \in A$ such that $x \mid 7$ (except itself)
$2 \mid 8$
$3 \mid 9$
$2 \mid 10$
$\not \exists x \in A$ such that $x \mid 11$ (except itself)
$2 \mid 12$
$\not \exists x \in A$ such that $x \mid 13$ (except itself)

Thus the maxima are the primes in $A$ -- which makes sense, for the divisibility relation.
We can similarly go through the list again and eliminate candidates for minima:

$2 \mid 4$
$3 \mid 6$
$4 \mid 8$
$5 \mid 10$
$6 \mid 12$
$7,8,\cdots,13$ only divide themselves in $A$

Thus the maxima are $7, 8,9,10,11,12$.

Note that there is no singular &quot;greatest maximum&quot;/&quot;least minimum&quot; for this relation. If there were, we would not have multiple maximal/minimal elements.
$2|4$ and no element $x\in A$ other than $2$ divides $2$, thus $ 2$ is the minimum of $ A$ according to the relation $ R$.

no element of $A$ divides $ 13$ but $2|12 $ and $12$ does not divide any other element of $A$. its follows that $ 12$ is the maximum of $A$.