A set $S$ is called upward directed set if it's reflexive and transitive, and for every two elements $a,b \in S$ there exist a $c \in S$ such that $c$ is an upper bound for both $a$ and $b$,e.g. $a\leq c$ and $b \leq c$.
The definition of a downward directed set is analogous, and a set which is both upward directed set and downward directed set is called directed set or filtered set.
Wikipedia states that:
Directed sets are a generalization of nonempty totally ordered sets. That is, all totally ordered sets are directed sets.
By definition every totally ordered set is reflexive and transitive and every two elements are related with respect to the relation defined on the set, but how it can be shown that every two pairs of elements are bounded above/below?
Also can someone give me an example of a filtered set that is not a totally ordered set?
The statement is not true for partially ordered sets, since for example if we consider a set $P=\left\{1,2,3\right\}$ and define a partially ordered set $S$ such that:
$(S,⊆):=\left\{\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{1,2\right\},\left\{2,3\right\}\right\}$
There deos not exist any upper bound for $\left\{1,2\right\}$ and $\left\{2,3\right\}$, although they both are bounded below by $\left\{2\right\}$.
There exists another definition for filtered set (upward directed set) that states:
A directed set is a set $S$ with a preorder such that every finite subset of $S$ has an upper bound.
The abonnement between these two definitions is that they both are preorder, but the first one states that every two elements are bounded above, and the second one states that every finite subset of $S$ is bounded above.
Clearly if let the two elements to be in a set then the set will be a finite set since it's cardinality is a finite value, and also the set containing the two elements will be bounded above, since each elements contained in that set is bounded above, but I think the second definition give us a more generalized concept since the subset may have more than two elements and still bounded.
Finally can someone prove the theorems in the two yellow boxes and verify my examples?
Theorem. Let $S$ be a set endowed with a preorder "$\le$". Then each finite subset $F$ of $S$ has an upper bound if and only if each two-element subset $F$ of $S$ has an upper bound.
Proof (with details). ($\Rightarrow$) This is trivial, because each two-element subset $F$ of $S$ is finite, so it has an upper bound.
($\Leftarrow$) We prove the following claim “each finite subset $F$ of $S$ has an upper bound” by an induction with respect to a number $|F|$ of elements of the set $F$. If $|F|=1$ then $F=\{x\}$ for some $x\in S$ so $x$ is an upper bound for each (that is, for a unique) element of $S$. If $|F|=2$ then the claim holds because each two-element subset $F$ of $S$ has an upper bound. Assume that the claim if proved for some natural number $n$ and let $F$ be a subset of $S$ such that $|F|=n+1$. Let $\{x_1,\dots, x_{n+1}\}$ be the elements of $F$. Put $F’=\{x_1,\dots, x_n\}$. Since $|F’|=n$, by the inductive hypothesis the elements of the set $F’$ have a common upper bound, say $x\in S$. The set $G=\{x,x_{n+1}\}$ consists of two elements, if $x\ne x_{n+1}$ and of one element, if $x=x_{n+1}$. So $|G|=1$ or $|G|=2$. By the cases considered in the beginning of the induction, all elements of $G$ have a common upper bound, say $y$. We claim that $y$ is a common upper bound for elements of $F$. Indeed, let $z$ be any element of $F$. Only the following two cases are possible.
We have $z=x_{n+1}$. Then $y$ is an upper bound for $z$ since $y$ is an upper bound for $x_{n+1}$.
We have $z\ne x_{n+1}$. Then, since $z\in F$, we have $z\in F’$. Then $z\le x$, since $x$ is a common upper bound for elements of $F’$. Since $x\le y$, by the transitivity of the relation “$\le$”, we have $z\le y$, so $y$ is an upper bound for $z$.
Thus, anyway, $y$ is an upper bound for $z$, which follows that $y$ is a common upper bound for elements of $F$. This finishes the step of the induction, and, therefore, the proof of the theorem. $\square$