Show that $F\cup\{a\wedge x: x\in F\}$ is a filter if $F$ is a filter

Question

Let $(L\leq)$ be a distributive lattice. Let $F$ be a filter on $L$. If $a\in L\setminus F$, show that $G:=F\cup\{a\wedge x:x\in F\}$ is also a filter in $L$

Now $G\neq\emptyset$ as $F\subset G$, and it is quite easy to see that $G$ is downward directed (i.e. closed under $\wedge$). But I am having trouble showing that $G$ is upwardly closed, i.e. if $z\in G$ and $y\geq z$ then $y\in G$. It is obvious if $z\in F$ but if $z=a\wedge x$ for some $x\in F$ then I am having trouble seeing why $y\geq a\wedge x\Rightarrow y\in G$. My only ideas where to somehow use $y\vee x$ as $F$ being upwardly closed we have $y\vee x\in F$ (as $x\in F$); but I don't get very far and not sure this on the right path so any help and assistance is greatly needed.

Thanks in advance.

Answer 1

$ab = a \inf b$; $a + b = a \sup b$.
If $x \in F$ and $ax <= y$, then $ax <= ay$;
$ay = ax + ay = a(x + y)$ and $x + y \in F$.

Answer 2

It is not true in general. Here is a simple example.

Every chain is a distributive lattice. Consider a three-element chain $\{0, 1, 2\}$ with $0 < 1 < 2$ and let $F = \{2\}, a = 0$. Then $G = \{2\} \cup \{0 \wedge 2\} = \{0, 2\}$ which is not an upward closed set since $0 \in G,$ but $0 < 1$ and $1 \notin G$.