ACC implies that every ideal is principal

Question

Let $(L,\leq)$ be a lattice such that every ascending chain of elements in $L$ is stationary. A lattice-ideal $I$ in $L$ is called principal if there exists $x \in I$ such that $I=\downarrow x= \lbrace y \in L\mid y \leq x \rbrace$. How does one show that every ideal in $L$ is principal? It is clear if $I$ has countable many elements: set $I=\lbrace x_i \mid i \in \mathbb{N} \rbrace$ and consider the chain $x_0 \leq x_0 \vee x_1 \leq \vee x_0 \vee x_1 \vee x_2 \vee\leq ...$

But I am not sure about the uncountable case. Is there a more general approach?

Answer 1

Suppose $I$ is an ideal that is not principal. Pick an element $x_0\in I$.* Since $x_0$ does not generate $I$, there is some $x_1\in I$ such that $x_1\not\leq x_0$. Then since $x_0\vee x_1$ does not generate $I$, there is some $x_2\in I$ such that $x_2\not\leq x_0\vee x_1$. Continuing this process recursively, we get a sequence of elements of $I$ such that $$x_0<x_0\vee x_1<x_0\vee x_1\vee x_2<\dots,$$ contradicting the ACC.

_{*Note that for this result to be true, your definition of "lattice" should include being bounded, or else you need to require $I$ to be nonempty.}