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The following is an example given by my professor, but there is an equality that I don't understand.

Let a partially ordered set $(S,\preceq)$ is the union of three sets such that $S=X\cup Y\cup Z$ such that $x\preceq y$ and $x\preceq z$ for all $x\in X, y\in Y, z\in Z$, but no $y\in Y$ and no $z\in Z$ are comparable.

Now let $T=S\cup$ {$\nu,\eta$} and $T'=S\cup$ {$\nu$} where

$x\prec\nu\prec y$ for all $x\in X$ and all $y\in Y$, $x\prec\eta\prec z$ for all $x\in X$ and all $z\in Z$, but $\nu$ and $\eta$ are not comparable.

It continues to say that $L(Y)=(\nu]\cap S=X=L(Z)$. However, I don't see how the first equality holds. Since $\nu=\inf_{T'}Y$, it would follow that $\nu\in L(Y)$, but isn't $\nu\notin (\nu]\cap S$, since $\nu\notin S$?

Perhaps I have mistook $L(Y)$ as $L_{T'}(Y)$, and so it is actually not the case that $\nu\in L(Y)$? Thanks for any answer to my doubts.

(Also, is there a different way to TeX { and } on this site? { and \left{ don't seem to translate.)

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    Try \\{ for TeX {2010-09-13
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    \lbrace and \rbrace also work.2010-10-14
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    Sometimes the braces seem to want one backslash for me, sometimes two. I just see what works.2011-01-12

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