$P$ 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 $a \nleqslant y$

Solution 1:

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.