Suprema always exist $\iff$ Infima always exist

Question

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: $$\left(\forall A\subseteq X: \sup A\text{ exists}\right)\iff\left(\forall A\subseteq X: \inf A\text{ exists}\right)$$

Answer 1

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: $$\left(\sup A_-=\inf A\right)\text{ resp. }\left(\inf A_+=\sup A\right)$$

Answer 2

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 $i = \inf Z$.

Answer 3

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.

Answer 4

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!