Recall: A set $X$ together with a binary relation $\leq$ such that for all $x,y,z\in X$,
O$1$. $x\leq x$
$O2$. $x\leq y$ and $y\leq x$ then $x=y$
$O3$. $x\leq y$ and $y\leq z$ then $x\leq z$
is called an ordered or sometimes a partially ordered.
- An ordered $(X,\leq)$ is called a total order if for all $x,y\in X$, either $x\leq y$ or $y\leq x$.
Theorem 1. Show that total orders are max orders.
Let $\leq$ be a total order on $X$. There is no partial order on $X$ which extends $\leq$ because $\leq$ is total order.
Theorem 2. Show that maximal order is total order.
Let $\leq$ be an partial order on $X$ which is maximal. Since $\leq$ is maximal, any two elements of $X$ can be compatible via $\leq$, hence $\leq$ is total order.
May you check my proofs? Can you give me alternative proof? Thanks...
Let S be a linear order. Assume a <= b is not in the order of S.
Define S' with the order including the order of S and a <= b.
As S is linear, b <= a is in S.
Thus within S', a <= b and b <= a; a = b.
As a <= a is in the order of S and a = b, a <= b is is in the order of S.
That contradiction shows the impossibility of extending the order of S.
Let S be maximally ordered.
Assume some a,b with not a <= b and not b <= a.
Show that the order of S can be extended to include a <= b.
As S is maximally ordered, that is a contradiction.
Thus the negation of the assumption: a <= b or b <= a.