Ian Charlesworth

Suppose that $(S, \preceq)$ is an ordered set, that $E \subseteq S$, and that $t$ is a lower bound for $E$. Then for any $x \prec t$, $x$ is a lower bound for $E$.
1. True.
2. True, but only with the additional assumption that $t = \inf E$.
3. False.
Suppose that $(S, \preceq)$ is an ordered set, and $E \subseteq S$ is bounded above. Then $E$ has infinitely many distinct upper bounds...
1. ...with no further assumptions.
2. ...provided that $E$ does not have a maximum element.
3. ...provided that $S$ does not have a maximum element.
4. ...if $S$ has the least upper bound property.
Suppose $(S, \square)$ is an ordered set, and $E \subseteq S$. We say $E$ is bounded above if...
1. ...for every $t \in E$ there is some $s \in S$ with $t \square s$.
2. ...$\sup E$ exists.
3. ...there is some $s \in S$ so that for every $t \in E$, $t \square s$.
4. ...whenever $s \in S$ and $t \in E$, we have $t \square s$.