If $B$ if a finite Boolean algebra, then it contains an atom Let $B=\{a_1,\dots,a_n,0,1\}$ be a finite Boolean algebra. I want to show that there exists an atom $x\in B$. So I want to show that there exists $x\in B$, such that for each $a\in B$ for which $a<x$, we have $a=0$.
I was thinking of maybe looking at $a=\bigwedge\{a_1,\dots,a_n\}$. If $a\neq 0$, then it seems to me that $a$ is an atom. So assume $a=0$. Is it possible to shrink the size of $\{a_1,\dots,a_n\}$ so that we get $a\neq 0$? And for the biggest set for which $a> 0$ we could argue that $a$ is an atom?
(Short remark: I do know that every finite Boolean algebra equals $\mathcal P(\operatorname{At}(B))$, but to me it seems that we use the fact that $\operatorname{At}(B)$ is nonempty, or at least, that seems to be assumed. So I want to prove this from scratch.)
 A: Suppose there is no atom. Then there exists $i_1$ such that $a_{i_1}<a_1$, and $a_{i_2}<a_{i_1}$, etc. Since there are only finitely many $a_i$'s, $i_k=i_{k'}$ for some $k\ne k'$. Thus, $a_{i_k}<a_{i_k}$ by transitivity, which is impossible.
A: A better way would be to note that for every $a$ and set $\{a_0, \ldots,a_n\}$ with $x=\wedge \{a_0, \ldots, a_n\}>0$ we have either, $x \wedge a >0$ or $x\wedge (1-a) >0$. Using this and the finiteness of $B$ you can recursively build a finite subset with non-zero meet $x$ such that for every $a$, exactly one of the alternatives $x\le a$ or $x\le 1-a$ hold. 
How this works: Let $B=\{0,1,a_0,a_1,\ldots, a_N\}$.
Then first, we let $H_0=\{1\}$, and note that $x_0 =\wedge H_0 >0$. 
Next, assume that for some $n\ge 0$ we have defined the finite set $H_{n} \subset B$, so that $x_{n} = \wedge H_{n} > 0$. If $n=N+1$ then we are finished. Otherwise, either $x_{n} \wedge a_{n} >0$ or $x_{n} \wedge (1-a_{n}) >0$. As such, we define $H_{n+1}$ by  $H_{n+1}= H_{n} \cup \{ a_{n}\}$ if $x_{n} \wedge a_{n} > 0$ and $H_{n+1}= H_{n} \cup \{ 1-a_{n}\}$ otherwise. Then $\wedge H_{n+1} > 0$ and we can continue.
At the end of this process you have $x = \wedge H_{N+1}>0$ and for every $a\in B$ either $x\le a$ or $x\le (1-a)$.
A: I found the answers provided quite hard or unclear, but I think I have an attempt at a proof myself:
Consider for each $x\in B\setminus\{0\}$ a "maximal decreasing chain". What I mean by a decreasing chain is a sequence $0<a_1<a_2<\dots<a_t<x$, where $a_i\in B$. Consider $y\in B$ such that is has a smallest maximal decreasing chain of all $x\in B$. Assume that this chain is greater than 2, which means that we have $0<a_1<\dots <a_t<x$, for $t\in \mathbb N_{\geq 1}$. Now consider $0<a_1$. This is the maximal decreasing chain for $a_1$, for if it were bigger, then we could have increased the maximal decreasing chain for $y$, which is a contradiction. So it must hold that the maximal decreasing chain for $y$ is $0<y$. Therefore, $y$ is an atom.
A: Assume that $x\in B$ is not an atom. Then there are $x_a,x_b\in B$ such that $x_a\vee x_b=x$, $x_a,x_b\neq x$, and $x_a,x_b\neq 0$. If $x_a$ is not an atom, then we can find $x_{aa},x_{ab}$ such that $x_{aa}\vee x_{ab}=x_a$, $x_{aa},x_{ab}\neq x_a$, and $x_{aa},x_{ab}\neq 0$. Continuing in this way we find $$x=x_1\vee x_2\vee ...\vee x_n$$ such that $x_i\neq0$ and, because $B$ is finite, if $x_i=x_a\vee x_b$ with $x_a,x_b\neq 0,x_i$, then $x_a,x_b\in\{x_1,...,x_n\}$. When this happens $n$ can be decreased maintaining the same property.
Take $n$ minimum with the property above. Each of the $x_i$ must be atoms, otherwise one can split them and reduce $n$ further.
