Playground

Problem 1: Linear Extension

Let $r\subseteq A\times A$ be an arbitrary relation and $\leq$ a partial order on $A$ .

a) Assume $a,b\in A$ to be incomparable: $a\not\leq b$ and $b\not\leq a$ . Show that the transitive closure of the relation $r_{\text{ex}}$ is a partial order on $A$ .

$r_{\text{ex}} = \{(x,y) | x\leq y\}\cup\{(a,b)\}$

b) Use (a) to show if $A$ is finite then every order $\leq$ on $A$ has a linear extension. A total order $\leq_{1}$ is a linear extension of a partial order $\leq$ if, whenever $x\leq y$ implies that $x\leq_{1} y$ .

c) Use (a) to show there is a finite number of total orders $(A,\leq_1),\cdots,(A,\leq_n)$ such that for all $a,b\in A$ we have

$a\leq b \Leftrightarrow (a\leq_1b \wedge\cdots\wedge a\leq_n b)$

d) Show that $\leq_1$ is a linear extension of $\leq$ if and only if $(A,\leq_1)$ is a total order and the identity function is a monotone function from $(A,\leq)$ to $(A,\leq_1)$ .