Is this an equivalent condition for algebraic posets? The definition of "algebraic poset" in Continuous Lattices and Domains, Definition I-4.2, says that, for all $x \in L$, 


the set $A(x) = {\downarrow} x \cap K(L)$ should be a directed set, and
$x = \bigsqcup ({\downarrow} x \cap K(L)$.


Here $L$ is a poset, $K(L)$ is the set of compact elements of $L$, and ${\downarrow} x$ means $\{y \mid y \sqsubseteq x\}$.

I was a bit surprised by the first condition.  It is an easy argument to show that, if $k_1$ and $k_2$ are in $A(x)$ then $k_1 \sqcup k_2$ is also in $A(x)$.  So, all nonempty finite subsets of $A(x)$ have upper bounds in it.  The only question is whether the empty subset has an upper bound in it, i.e., whether $A(x)$ is nonempty in the first place.  So,


Is it ok to replace the first condition with $A(x)$ is nonempty?
What is an example of a situation where $A(x)$ is empty?




Note added: How is $k_1 \sqcup k_2$ in A(x)?  First, since $k_1 \sqsubseteq x$ and $k_2 \sqsubseteq x$, we have $k_1 \sqcup k_2 \sqsubseteq x$.  Second, $k_1$ and $k_2$ are compact.  So, any directed set that goes "beyond" them must "pass" them.  Suppose a directed set $u$ also goes beyond $k_1 \sqcup k_2$, i.e., $k_1 \sqcup k_2 \sqsubseteq \bigsqcup u$.  Since it has gone beyond $k_1$ and $k_2$, it must have passed them, i.e., there are elements $y_1, y_2 \in u$ such that $k_1 \sqsubseteq y_1$ and $k_2 \sqsubseteq y_2$.  Since $u$ is a directed set, it must have an upper bound for $y_1$ and $y_2$, say $y$.  Now, $k_1 \sqcup k_2 \sqsubseteq y \in d$.  This shows that $k_1 \sqcup k_2$ is compact.  The two pieces together say $k_1 \sqcup k_2 \in A(x)$.
An example where $A(x)$ is empty is the set of real numbers $\mathbb{R}$ with the usual ordering. It has no compact elements at all.

If we assume the second condition then $A(x)$ cannot be empty: if $A(x) = \emptyset$ then by the second condition $x$ is the empty join, therefore the least element of $L$, which is compact, therefore $x \in A(x) = \emptyset$, a contradiction.

Your proposal to replace the first condition with non-emptyness does not work. Consider the poset $L$ which consists of two copies of $\mathbb{N}$ and $\infty$, where we write $\iota_1(n)$ and $\iota_2(n)$ for the two copies of $n$, ordered by:


$\iota_1(m) \leq \iota_1(n) \iff m \leq n$
$\iota_2(m) \leq \iota_2(n) \iff m \leq n$
$x \leq \infty$ for all $x$.


In words, we have two incomparable chains with a common supremum. All elements are compact except $\infty$. Now:


${\downarrow}x \cap K(L) \neq \emptyset$, obviously.
$x = \bigvee ({\downarrow}x \cap K(L))$, obviously.
The set ${\downarrow}\infty \cap K(L) = \mathbb{N} + \mathbb{N}$ is not directed.