# Can one show that $L(E)=\langle s]$ on a partially ordered set?

This is a follow up question to one I posted earlier. I'm trying to decide that if for $(S,\preceq)$ a partially ordered set and $E\subseteq S$, one has $L(E)=\langle s]$ for some $s\in S$ iff $\inf E$ exists, and in particular, $L(E)=\langle\inf E]$.

I'm simply curious about showing that $L(E)\subseteq\langle\inf E]$ when $\inf E$ exists. Taking some $x\in L(E)$, if $x$ is comparable to $\inf E$, then $x\preceq\inf E$, and so $x\in\langle\inf E]$. But what if $x$ and $\inf E$ are not comparable? Is it still possible to show such an inclusion?

Edit: $\langle s]$ is the set {$x\in S \ | \ x\preceq s$}.

-
Sorry, what does the notation $\langle s]$ mean? – Jonas Meyer Sep 8 '10 at 6:25
My apologies for not making it clear. I have put the explanation in the original question. – yunone Sep 8 '10 at 6:31
Good, that is what I guessed. Thanks for clarifying. – Jonas Meyer Sep 8 '10 at 6:32

If I correctly understand your question, it boils down to whether $\inf E$ is "larger" than or equal to every lower bound of $E$. Isn't this the definition? A greatest lower bound of a set must be, by definition, comparable to and "larger" than (or equal to) every lower bound.
So the answer is yes: If $E$ has an inf, then $L(E)=\langle \inf E]$. Anything smaller than a lower bound is also a lower bound, which gives the right to left inclusion. The other inclusion follows from the fact that $\inf E$ is the "greatest" lower bound.