63 lines
2.8 KiB
TeX
63 lines
2.8 KiB
TeX
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\begin{definitionOf}{chain}
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Given a \Poset $\langle \LL, \lte \rangle$, a {\em chain} is a (possibly empty) set $C \subseteq \LL$ that is totally ordered, that is,
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for all $x, y \in C$ either $x \lte y$ or $y \lte x$.
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\end{definitionOf}
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\begin{definitionOf}{setordering}
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Given a preorder, $\langle \LL, \lte \rangle$
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We define a set ordering $\langle \powersetO(\LL), \lte \rangle$
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s.t. for any $X, Y \subseteq \LL$
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\begin{align*}
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X \lte Y \textrm{ iff } (X \upclosure) \supseteq Y
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\end{align*}
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\end{definitionOf}
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\begin{definitionOf}{chaincompleteposet}
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We call \Poset $\langle \LL, \lte \rangle$ {\em chain-complete} if $\lub \LL$ is unique and in $\LL$ and for ever chain $C \subseteq \LL$
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\begin{enumerate}
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\item $\lub \LL$ is unique and in $\LL$, and
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\item for every chain $C \subseteq \LL$, we have $\lub C \in \LL$
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\end{enumerate}
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\end{definitionOf}
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\begin{definitionOf}{antisymmequiv}
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Given a set $\powersetO(\LL)^2$, define $\antisymmequiv{\powersetO(\LL)^2}$ to be the set obtained by removing any elements $x \in \powersetO(\LL)^2$ s.t. $x \not= x \upprecision$.
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\end{definitionOf}
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\begin{remark}
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The \Preorder $\langle \powersetO(\LL)^2, \lteSetLPrecision \rangle$ lacks \Antisymmetry, thus it is not a \Poset.
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We can construct and equivalence relation $\equiv$ based on the \Antisymmetry condition:
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$x \equiv x' \textrm{ iff } x \lteSetLPrecision x' \textrm{ and } x' \lteSetLPrecision x$.
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Each equivalence class $[x]$ can be uniquely represented by $x \upprecision$. Thus, another way of viewing
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$\langle \antisymmequiv{\powersetO(\LL)^2}, \lteSetLPrecision \rangle$
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is as the quotient under the above equivalence relation.
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\end{remark}
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\begin{lemma}
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Given $\langle \antisymmequiv{\powersetO(\LL)^2}, \lteSetLPrecision \rangle$, and $x, y \in \powersetO(\LL)^2$
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\begin{enumerate}
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\item $x \lteSetRPrecision y$.
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\item if $x \lteSetLPrecision y$ then $x \lteSetLRPrecision y$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1) We have $\top_{\ltePrecision} \in y$. (2) Follows from (1).
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\end{proof}
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\begin{propositionOf}{chaincompletepreserve}
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Let $\langle \LL, \lte \rangle$ be a \CompleteLattice and $o: \LL^2 \rightarrow \powersetO(\LL)$ \AnNdao.
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The structure $\langle \antisymmequiv{o\image{R}}, \lteSetL \rangle$ is a \ChainCompletePoset where $R$ is one of the following:
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\begin{enumerate}
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\item\label{chaincompletepreserve:enum:a} $\LL^2$,
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\item\label{chaincompletepreserve:enum:b} $\LLc$
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\item\label{chaincompletepreserve:enum:c} $\postfixpointsOf(o)$
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\item\label{chaincompletepreserve:enum:d} $\prefixpointsOf(S(o))$
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\end{enumerate}
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\end{propositionOf}
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\begin{proof}
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This should be simpler, the incoming is chain complete poset so should inject one ?
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\end{proof} |