Inverse Image under Order Embedding of Strict Upper Closure of Image of Point
Theorem
Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets.
Let $\phi: S \to T$ be an order embedding of $\struct {S, \preceq}$ into $\struct {T, \preceq'}$
Let $p \in S$.
Then:
- $\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
where $\cdot^\succ$ and $\cdot^{\succ'}$ represent strict upper closure with respect to $\preceq$ and $\preceq'$, respectively.
Proof
Let $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$.
By the definition of inverse image:
- $\map \phi x \in \map \phi p^{\succ'}$
By the definition of strict upper closure:
- $\map \phi p \prec' \map \phi x$
Since $\phi$ is an order embedding:
- $p \prec x$
Thus by the definition of strict upper closure:
- $x \in p^\succ$
and so:
- $\map {\phi^{-1} } {\map \phi p^{\succ'} } \subseteq p^\succ$
Let $x \in p^\succ$.
By the definition of strict upper closure:
- $p \prec x$
Since $\phi$ is an order embedding:
- $\map \phi p \prec' \map \phi x$
Thus by the definition of strict upper closure:
- $\map \phi x \in \map \phi p^{\succ'}$
Thus by the definition of inverse image:
- $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$
and so:
- $p^\succ \subseteq \map {\phi^{-1} } {\map \phi p^{\succ'} }$
Thus by definition of set equality:
- $\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
$\blacksquare$