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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.

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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.

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  • $\begingroup$ Okay thanks. But why does the question say 'the lower bound of a set might not be unique' This implies to me there exists a set which has only one lower bound. $\endgroup$ – CAF Mar 6 '13 at 10:08
  • $\begingroup$ that is not carefully worded indeed. Any bounded below subset of the reals has infinitely many lower bounds. $\endgroup$ – Ittay Weiss Mar 6 '13 at 10:09
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Suppose that M, M ′ are suprema of A. Then M ≤ M ′ since M ′ is an upper bound of A and M is a least upper bound; similarly, M ′ ≤ M, so M = M ′. If m, m′ are infima of A, then m ≥ m′ since m′ is a lower bound of A and m is a greatest lower bound; similarly, m′ ≥ m, so m = m′. If inf A and sup A exist, then A is nonempty. Choose x ∈ A, Then inf A ≤ x ≤ sup A since inf A is a lower bound of A and sup A is an upper bound. It follows that inf A ≤ sup A.

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  • $\begingroup$ This doesn't really address the OP's question, who wanted advice on the proof given in the post. $\endgroup$ – Kevin Long Feb 23 '17 at 18:26

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