# Suprema always exist $\iff$ Infima always exist

Let $X$ be a poset (or only preordered or even just equipped with a plain relation).

Is it true that suprema always exist iff infima always exist: $$\left(\forall A\subseteq X: \sup A\text{ exists}\right)\iff\left(\forall A\subseteq X: \inf A\text{ exists}\right)$$

• question corrected... – C-Star-W-Star Jun 22 '14 at 23:08
• If you are planning to answer your own question immediately, the least you can do is to say so explicitly in the question, so that other people can judge whether to spend time on it. But in general this should be something you do quite rarely; you have answered a large number of your own questions recently. – Carl Mummert Jun 23 '14 at 11:16
• Oh yes you're right it'd be nice if I mention that this is meant as Q&A - will do it next times. Thanks for the advice! – C-Star-W-Star Jun 23 '14 at 11:37
• Isn't $\mathbb{N}$ a counterexample to this, since it has no supremum but any subset has an infemum? Are people answering a slightly different question than what is asked? (i.e. suprema always exist for sets with an upper bound iff infema always exist for sets with a lower bound) – Carl Jun 23 '14 at 19:29
• @Carl: No, the natural numbers are not a counterexample. Consider the empty set. It has no infimum since the collection of lower bounds is the natural numbers itself and this one has no maximum. – C-Star-W-Star Jun 24 '14 at 11:09

Yes, indeed $\inf A=\sup A_-$ and $\sup A=\inf A_+$ with $A_+:=\{c:c\geq A\}$ and $A_-:=\{c:c\leq A\}$.

Define for brevity: $$A\leq c:\iff a\leq c\text{ for all }a\in A$$ Since suprema are the least upper bounds resp. infima the greatest upper bounds it holds: $$\left(\sup A_-=\min A_{-+}\leq A_{-+}\right)\text{ resp. }\left(\inf A_+=\max A_{+-}\geq A_{+-}\right)$$ But $A\subseteq A_{-+}$ resp. $A\subseteq A_{+-}$ so it also holds: $$\left(\sup A_-=\min A_{-+}\leq A\right)\text{ resp. }\left(\inf A_+=\max A_{+-}\geq A\right)$$ Since suprema are lower bounds resp. infima upper bounds it holds: $$\left(\sup A_-=\min A_{-+}\geq A_-\right)\text{ resp. }\left(\inf A_+=\max A_{+-}\leq A_+\right)$$ Concluding: $$\left(\sup A_-=\inf A\right)\text{ resp. }\left(\inf A_+=\sup A\right)$$

Hint: Let $(X, \le)$ be a partial order, where every subset $Y \subseteq X$ has a supremum. Given any, $Z \subseteq X$ - we want to show it has an infimum -, let $$i = \sup\{y \in X \mid y \le z, \quad \text{all } z \in Z\}$$ that is the sup of $Z$'s lower bounds. Now show $i = \inf Z$.

Suppose every subset of this poset has a supremum.

Given a set $A$, let $f(A)$ be the set of all lower bounds of $A$. Then $f(A)$ has a supremum. Every member of $A$ is an upper bound of $f(A)$, since every member of $A$ is $\ge$ every lower bound of $A$. Hence $\sup A\le$ every member of $A$. In other words, $\sup f(A)$ is a lower bound of $A$. And $\sup f(A)\ge$ every other lower bound of $A$ since $\sup A$ is by definition the supremum of the set of all lower bounds of $A$. Hence $\sup f(A)=\inf A$.

In other words, every subset of the poset has an infimum.

PS: We can strengthen the conclusion a bit, so that it applies to some posets in which not every set has a supremum. The strengthened conclusion says: If $A$ is any subset of a poset, and the set of all lower bounds of $A$ has a supremum, then $A$ has an infiumum.

First recall the pointwise order: for sets $A,B$, $$A \leq B :\equiv (\forall a \in A, b \in B :: a \leq b)$$ if either is a singleton set $\{x\}$ we write just $x$ for brevity; eg, we define upper and lower bounds as $$A_+ := \{ u \,:\, A \leq u \} \;\;,\;\; A_- := \{l \,:\, l \leq A\}$$

Anyhow, to the task at hand, $\;\;\; \text{sups exist} . \\ \equiv \text{for every set } A \text{ there is an element } s \text{ being the least upper bound of } A \\\equiv \forall A :: \exists s :: s \text{ upper bound of A and least such} \\\equiv \forall A :: \exists s :: \; A \leq s \leq A_+ \\\Rightarrow \forall A :: \exists s :: \; A_- \leq s \leq A_{-+} \; \text{by instantiation: it holds for all$A$, and so holds for$A_-$} \\\Rightarrow \forall A :: \exists s :: \; A_- \leq s \leq A \; \text{ since$A \subseteq A_{-+}$and$X \subseteq Y \geq l \Rightarrow l \leq X$} \\ \equiv \text{for every set } A \text{ there is an element } s \text{ being a lower bound of } A \text{ and the greatest such } \\\equiv \text{infs exist.}$

Hence, existence of sups implies existence of infs. The converse holds by duality, (or an exercise to the reader.)

This' essentially Freeze_S's answer in a more linear form; aiming for clarity.

Hope this helps!