Let $L$ be a lattice. We say that $a\in L$ is irreducible if for every $b,c\in L$ such that $a=b\vee c$ we can conclude that $a=b$ or $a=c$. If $L$ is a finite lattice prove that every element $a\in L$ can be written as $a=a_1\vee a_2 \vee \cdots \vee a_n$ where $a_1, a_2, ..., a_n$ are irreducibles.
How can I attack this problem?
Thanks!
Assume that $L$ satisfies the descending chain condition ($L$ does not possess any infinite descending chains). Notice that any finite lattice satisfies this condition.
Let $a \in L$. If $a$ is join-irreducible we are done. Otherwise it can be represented as $a = a_1 \vee a_2$, for some $a_1, a_2 \in L$ and also $a_1 \neq a, a_2 \neq a$, hence $a_1 < a, a_2 < a$. Either both of $a_1$ and $a_2$ are join-irreducible or one of them can be represented in the same form as above. If $a$ can't be represented in the desired form, this process will give you infinite descending chain, which contradicts the assumption. Hence this process ends after finitely many steps and $a = a_1 \vee \dots \vee a_k$ for some join-irreducible elements $a_1, \dots, a_k \in L$.