# Does the existence of an infimum imply that the set of lower bounds of a set is totally ordered?

Say we have a partially ordered set $(S,\preceq)$, and some subset $E\subseteq S$ such that $E$ is bounded below and $\inf E$ exists. My question is, since $S$ is not totally ordered is it possible to have two lower bounds that are not comparable, and hence be unable to say whether one lower bound is really the greatest lower bound since you cannot compare to every other lower bound?

If such a thing is not possible, (which I believe is the case since it is given $\inf E$ exists) would it also be fair to say that the existence of $\inf E$ implies that the set $L(E)$ of lower bounds of $E$ is a totally ordered subset of $S$?

Thanks.

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No, because even if a partially ordered set is not totally ordered, there can be a "largest" element. For a simple example, take a set $X$ with at least 2 elements, let $S$ be the power set of $X$ and let the partial order be inclusion. Then $\inf\{X\}=X$, but the set of lower bounds of $\{X\}$ is $S$, which is not totally ordered because $X$ has more than one element.