$P$ poset. $x = \bigvee(\downarrow x\cap U)\Rightarrow \forall x, y \in P$, with $y \lt x$, $\exists a\in U$ s.t. $a \le x $ and $a \nleqslant y$ Le $P$ be a partially ordered set, $U \subseteq P$ and $\downarrow x = \{y \in P : y \le x\}$ (a down set).
Show that if $\,\,\forall \,\,x \in P\,\,$ we have $\,\,x = \bigvee(\downarrow x\cap U)\,\,\Rightarrow\,\,\forall x, y \in P$, with $y \lt x$, then $\exists \,\,a\in U\,\,$ s.t. $\,\,a \le x \,\,$ and $a \nleqslant y$. Prove also that if $P$ is a complete lattice then also $\Leftarrow $ is true.
When $P$ is a finite lattice everything works fine, in fact: let $L$ be a finite lattice and $x, y \in L\,\,$ s.t.$x \nleq y$.  Let be  $S := \{z \in L : z \le x\ , z \nleq y \}$, since $L$ is finite and $x \in S$ $\,\,$I can pick $a \in S$ which is minimal in $S$. $\,\,$ If $\,\,\,a = c \vee d$ for some $c, d,\in L$, with $c<a$ and $d <a$, then $c, d \notin S$ because of the minimality of $a$, thus: $c, d \le y$ with $c<a\le x$ and $d<a\le x \Rightarrow a = c\vee d \le y$, a contraddiction. So $a$ has to be join-irreducible $\Rightarrow a \in J(L)\,\,\,\,$ (actually $\,\,a = \bigvee\{z \in J(L) : z \le a \} = \bigvee(\downarrow a \cap J(L)\,\,\,)$
but how to prove it for a poset without hypothesis on its cardinality?
 A: We prove the left to right direction first ($\Rightarrow$). So let $x < y \in P$. We aim for a proof by contradiction, so suppose that for all $a \leq x$ with $a \in U$ we also have that $a \leq y$. This is exactly saying that for all $a \in \downarrow x \cap U$ we have $a \leq y$. That means that $x = \bigvee (\downarrow x \cap U) \leq y < x$, a contradiction. So there must be some $a \leq x$ with $a \in U$ such that $a \not \leq y$, as required.
For the other direction ($\Leftarrow$), we assume $P$ is a complete lattice. Let $x \in P$ be arbitrary. Since $P$ is a complete lattice, $\bigvee (\downarrow x \cap U)$ exists and we can set $y = \bigvee (\downarrow x \cap U)$. We need to show that $x = y$. Clearly $y \leq x$, as $x$ is an upper bound of $\downarrow x \cap U$. Again aiming for a contradiction, we assume that $x \neq y$ and hence $y < x$. By assumption we can find $a \in U$ with $a \leq x$ such that $a \not \leq y$. However, this means precisely that $a \in \downarrow x \cap U$ and $a \not \leq y$. So $y$ is not an upper bound of $\downarrow x \cap U$, contradiction. We conclude that $x = y$, as required.
