Prove that a lower bound of a set might not be unique but the infimum of a given set is unique.
Attempt: Consider some $E \subset \mathbb{R} $ such that $E \neq \emptyset$. $E$ is bounded below $\iff \exists m \in \mathbb{R}\,\, \text{such that}\,\, a \geq m\,\, \forall a \in E $.
Presumably the only case where a lower bound need be unique is if the end interval of the set is infinity i.e some subset of the form $(-\infty, p)$, where $p \in \mathbb{R}$.
To prove the uniqueness of infimum: Suppose there exists two such infimum $t_1, t_2 \in \mathbb{R} $ such that $t_1 \leq a $ and $ t_2 \leq a\,\forall a \in E$. So $t_1$ and $t_2$ are both lower bounds for the set $E$ and in particular, since $t_1$ and $t_2$ are both infimum, $t_1 \leq t_2$ and $t_2 \leq t_1$ by the definition of infimum. Hence by the Trichotomy Principle, we conclude $t_1 = t_2$.
Is this okay? If so, am I correct in what I said above that provided the subset does not extend to $-\infty$, lower bounds for the set need not be unique?
Many thanks.
Best Answer
Your proof is ok. The remark preceding it is not. A set of the form $(-\infty , p)$ is simply not bounded below. In general, if a set $S\subseteq \mathbb R$ is bounded below that it has infinitely many lower bounds, but only one greatest lower bound.
Note: the plural of infimum is infima.