I've already showed that $L$ is a complete lattice. I need to show it is non compactly generated, so not every element of $L$ is supremum of compact elements.
I see that only $0$ is compact element in that lattice, it's obvious, but how I should show that? Have no idea how do it properly.
I think that after showing this I just need to show $0$ is his own supremum and no other element of $L$ is supremum of compact elements. Am I right?
Waiting for any hints or suggestions. Thanks!
For any $x \in (0, 1]$ we have that $x$ is the supremum of $[0, x)$. The latter is a directed set. However, even though we have $x \leq \sup [0, x)$, there is no element $y \in [0, x)$ such that $x \leq y$. This shows that $x$ is not compact. So no nonzero element is compact.
Of course, $0$ is trivially compact. Since a directed set $D \subseteq [0, 1]$ is non-empty by definition, there must be some $d \in D$ and we always have $0 \leq d$.