bounded distributive lattice which is not pseudocomplemented

Question

Currently I'm studying lattice theory course using Lattice theory by gratzer... I am stuck at the question that "Give an example of bounded distributive lattice which is not pseudocomplemented" I know that every finite distributive lattice is pseudocomplemented so that particular lattice must be infinite Please answer

Answer 1

Let $X$ be an infinite set. Let $L$ be the lattice consisting of all finite subsets of $X$ along with one more set, namely the entire set $X$. This is a bounded distributive lattice. (The bounds are $\emptyset$ and $X$.)

$L$ is not pseudocomplemented, since, for $x\in X$, there is no largest finite subset of $X$ that is disjoint from $\{x\}$ and $X$ itself is not disjoint from $\{x\}$.