Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
Try \\{ for TeX { –  Asaf Karagila Sep 13 '10 at 20:37
\lbrace and \rbrace also work. –  Paul VanKoughnett Oct 14 '10 at 4:47
Sometimes the braces seem to want one backslash for me, sometimes two. I just see what works. –  Ross Millikan Jan 12 '11 at 5:00
add comment

1 Answer

I do not think you know that $\nu=\inf_{T'}(Y)$. Since $S\subset T'$, you know that if $\inf_{T'}(Y)$ and $\inf_S(Y)$ both exist, then $\inf_S(Y)\preceq\inf_{T'}(Y)$; but you could have equality (e.g., if $Y$ has a minimum). If you have equality, then $\nu$ is not the infimum of $Y$ in $T'$. You do know that $\nu\preceq\inf_{T'}(Y)$, because $\nu$ is a lower bound for $Y$ in $T'$, but the inequality could be strict. I would argue that you do not know that $L(Y)=(\nu]\cap S$, because it excludes the possibility that there is a $y\in Y$ that is a lower bound for $Y$. For the same reason, $L_{T'}(Y)$ cannot equal $(\nu]\cap S$.

So, presumably you are also assuming that $Y$ has no minimum. If that's the case, then I think that $L(Y)$ is meant to be $L_S(Y)$, rather than $L_{T'}(Y)$ or $L_T(Y)$.

P.S. I'm assuming that $L(A)$ means "lower bounds of $A$"; it's possible it is supposed to mean "strict lower bounds of $A$" (that is, $L(A) = \{x|x\prec y$ for all $y\in Y\}$), in which case my objection above vanishes. In any case, I suspect that $L(Y)$ is meant to be $L_S(Y)$, $S$ being the set that we started with so the one that it makes sense to take as understood from context.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.