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In my book, the author defines posets formally in the following way:

Let $P$ be a set, and let $\le$ be a relationship on $P$ so that,

$a$. $\le$ is reflective.

$b$. $\le$ is transitive.

$c$. $\le$ is antisymmetric.

Say for $a$, does this merely mean that if some element $x\in P$, $x$ should always have the same relation to itself? and for $b$ if $x$ has the relation to $y$ and $y$ has the relation to $z$, this implies that $x$ has the relation to $z$?

Moreover, when trying to determine if a something is a poset,do I just have to determine if such a relationship exists? And that relationship is not necessarily the usual meaning of "$\le$"

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Yes, to each of your questions. A poset is a set $P$ and a partial-order relation defined on the set; when trying to determine if a relation and the set on which it is defined is a poset, you need to check each of the defining properties of a partial order relation with respect to the set. That is true when testing any relationship, not always denoted by $\leq$.

(You need to also show that the relation is antisymmetric, if it is to be a poset.)

In the following, we use the notation "$\,\leq\,$" to represent any partial order relation, not just the narrower/usual meaning of "$\,\leq\,$" ('is less than or equal to'), which also defines a poset on suitable sets. There are many relations which, together with the sets on which they are respectively defined, form a poset. You'll likely encounter some such examples in you studies and/or in problem sets.

A poset consists a set $P$ and a relation on the set. A relation on the set form a poset if and only if:

  • $\forall a \in P, \;a\leq a\;$ (reflexity);
  • $\forall a, b, c \in P,\;$ whenever $a \leq b$ and $b \leq c$, then this implies $a \leq c\;$ (transitivity);
  • $\forall a, b \in P,\;$ whenever $a \leq b$ and $b\leq a$, then this implies $a = b\;$ (antisymmetry).

So given a relation and a set on which it is defined, in order to determine whether you have a poset, you need to confirm that each of these three properties hold for all $a, b, c$ in the set.

Put differently, if there are no $a \in P$ for which reflexivity fails, then the relation reflexive; if there are no $a, b, c \in P$ such that transitivity fails, then the relation is transitive. And if there are no $a, b\in P$ in the set such that antisymmetry fails, then the relation is antisymmetric.

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  • $\begingroup$ Thanks for the reassurance! $\endgroup$
    – MITjanitor
    Mar 10 '13 at 20:56
  • $\begingroup$ You're welcome! $\endgroup$
    – amWhy
    Mar 10 '13 at 20:59
  • $\begingroup$ Are you good with this now, Calc1DropOut? Any lingering questions? $\endgroup$
    – amWhy
    Mar 10 '13 at 21:32
  • $\begingroup$ Sorry, forgot to accept. I'm good now, thanks! $\endgroup$
    – MITjanitor
    Mar 10 '13 at 21:34
  • $\begingroup$ Great! good luck... $\endgroup$
    – amWhy
    Mar 10 '13 at 21:34
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Your statements are correct.

On your last question: A poset is a pair ($P$, $R$), where $P$ is a set and $R$ is a relation on $P$ (which must have properties a, b and c). So the relation is a part of the poset, there is no freedom to choose it.

Hence to determine if something is a poset, you don't have to determine if a suitable order relation exists (it always does), but if the set together with the given relation is a poset. The relation should always be clear from the context, in particular if the relation symbol is not ''$\leq$''.

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  • $\begingroup$ To emphasise the last point, one notes that $=$ always makes a set into a poset. $\endgroup$
    – Zhen Lin
    Mar 10 '13 at 21:15

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