In this link:
http://en.wikipedia.org/wiki/Szpilrajn_extension_theorem
in the section "lemma". The author construct a partial order $T$ on a set $X$ which extends another partial order $R$ in the following: ($R$ is a partial order which is not total) Take $x,y\in X$ such that $x\not R y$ and $x\not R y$, then define:
$Rx=\{z\in Z : zRx\}$
$yR=\{z\in Z : y Rz \}$
and $S=Rx\times yR\subset X\times X$. (I suppose that $R$ is reflexive). And finally $T=R\cup S$
He claim that this set is "trivially" a partial order which extends $R$. Actually $T$ trivially extends $R$ but, why $T$ is a partial order? In particular I don't know how to prove the transitivity, someone can show me why this property holds?
Thanks!
Proving transitivity of $T$: Let $aTb$ and $bTc$. We need to show that $aTc$.
Because $T = R \cup S$, we have four cases:
$1.$ $aRb$ and $bRc$
By transitivity of $R$, we have $aRc$, hence $aTc$.
$2.$ $aSb$ and $bSc$
$aSb \Rightarrow a\in \{x\} \cup Rx$. Also, $bSc \Rightarrow c\in \{y\} \cup yR$. Thus, $aSc$ and so $aTc$.
$3.$ $aRb$ and $bSc$
$bSc \Rightarrow b\in \{x\} \cup Rx$. So $b=x$ or $bRx$ and with $R$ being transitive, $aRb \Rightarrow aRx$. Thus $a\in Rx$. Also, $bSc \Rightarrow c\in \{y\} \cup yR$. Thus, $aSc$ and so $aTc$.
$4.$ $aSb$ and $bRc$
$aSb \Rightarrow b \in\{y\} \cup yR$. So $b=y$ or $yRb$ and with $R$ being transitive, $bRc \Rightarrow yRc$. Thus, $c\in yR$. Also, $aSb \Rightarrow a\in \{x\} \cup Rx$. Thus, $aSc$ and so $aTc$.