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:
$$\left(\forall A\subseteq X: \sup A\text{ exists}\right)\iff\left(\forall A\subseteq X: \inf A\text{ exists}\right)$$
 A: 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:
$$\left(\sup A_-=\inf A\right)\text{ resp. }\left(\inf A_+=\sup A\right)$$
A: 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 $i = \inf Z$.
A: 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.
A: 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!
