The dual of a poset $(S \leq)$ is the poset $(S, \geq)$, where for all $x,y \in S$ $x \leq y$ if and only if $ y \geq x$. A poset is self-dual if it is isomorphic to its dual.
- Verify that the dual of a poset is indeed a poset.
Given a poset $(S ,\leq)$, we can define its dual $(S, \geq)$
Using the definition $(x,y) \in S \leftrightarrow (y,x) \in R$,
we have $(R, \subseteq)$ and its dual is $(R, \supseteq)$
- Show that for any set $X$, the poset $(P(X) \subseteq)$ is self-dual.
Note: The $P(X)$ is the power set X
For any X, we have a poset isomorphism such that
$(P(X) ,\subseteq) \rightarrow ((P(X) ,\supseteq)$
$X \rightarrow X' = \alpha ' (x)$
If $X \subseteq Y \rightarrow X^c \supseteq Y^c$, then this map preserves the poset relation and it's an isomorphism. Therefore, $(X \subseteq)$ is self dual if it's isomorphic to its dual.
I'm not sure if this is correct or I should expand my proof further. It seems that I only have a part of it. Like for example on the first question.. wouldn't the last line be $(S, \subseteq)$ and it's dual is $(S \supseteq)$ because we are dealing with the poset $ (S \leq)$ and $(S \geq)$. There's a biconditional statement to begin with which produces two statements.
a. If the dual of a poset $(S \leq)$ is the poset $(S, \geq)$, where for all $x,y \in S$ $x \leq y$, then $ y \geq x$.
b. If $ y \geq x$, then the dual of a poset $(S \leq)$ is the poset $(S, \geq)$, where for all $x,y \in S$ $x \leq y$.
A poset is a partially ordered set if S is a set and R is a partial order on S -> that's the pair $(S,R)$
So the order relation R is a partial order if it's also reflexive, so wouldn't the reflexive definition apply to the first problems?