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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 {$x\in S \ | \ x\preceq s$}.

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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
up vote 3 down vote accepted

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.

I think the confusion arises because, as seen in your last question, having an inf does not imply that the set of lower bounds is totally ordered. It does imply, however, that all lower bounds are comparable to the inf.

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.

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Ah yes, that is definitely why I was confused. I knew an infimum had to be the 'largest' all other lower bounds, but I wasn't quite sure if that meant it would necessarily be comparable as I had never seen that mentioned explicitly. In retrospect, I feel a little silly. Anyway, thanks for answering both. – yunone Sep 8 '10 at 6:42
@yunone: You're welcome. I might have better answered the previous question by explicitly mentioning this. – Jonas Meyer Sep 8 '10 at 6:55

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