Ideal is Filter in Dual Ordered Set
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Theorem
Let $P = \struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then
- $X$ is ideal in $P$
- $X$ is filter in $P^{-1}$
where $P^{-1} = \struct {S, \succeq}$ denotes the dual of $P$.
Proof
Sufficient Condition
Let $X$ be ideal in $P$.
By definition of ideal in ordered set:
By definition of directed:
- $\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$
Then
- $\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$
By definition
- $X$ is filtered in $P^{-1}$
By definition of lower section:
- $\forall x \in X, y \in S: y \preceq x \implies y \in X$
Then
- $\forall x \in X, y \in S: x \succeq y \implies y \in X$
By definition
- $X$ is an upper section in $P^{-1}$.
Thus by definition of filter in ordered set
- $X$ is a filter in $P^{-1}$.
$\Box$
Necessary Condition
Let $X$ be a filter in $P^{-1}$.
By definition of filter in ordered set
By definition of filtered in $P^{-1}$:
- $\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$
Then
- $\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$
By definition:
- $X$ is directed.
By definition of upper section in $P^{-1}$:
- $\forall x \in X, y \in S: x \succeq y \implies y \in X$
Then:
- $\forall x \in X, y \in S: y \preceq x \implies y \in X$
By definition:
- $X$ is a lower section.
Thus by definition of ideal in ordered set:
- $X$ is an ideal in $P$.
$\blacksquare$
Sources
- Mizar article YELLOW_7:28
- Mizar article YELLOW_7:26