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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.)

Sha Vuklia
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  • Would you mind defining the expression $a < x$? Thanks! – Robert Lewis Mar 16 '18 at 17:50
  • Whatever you do, if you prove that a finite Boolean algebra has an atom, then you need a extra assumption which may be implicit in your reasoning, or it will be wrong: you need the assumption that $B$ is not trivial, that is, $|B|>1$. This is because there exist Boolean algebras with only one element, and those are, of course, atomless. – amrsa Mar 16 '18 at 18:44

4 Answers4

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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.

Eclipse Sun
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  • I first didn't realise that if $x>0$ is not an atom, then there must be a $y\in B$ such that $0<y<x$, because otherwise $z<x$ for $x\in B$ would imply $z=0$ (and then $x$ would be an atom). So now that I see that, this is definitely the easiest proof, thanks! – Sha Vuklia Mar 16 '18 at 19:00
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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 $x\le (1-a)$.

Not Mike
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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.

Sha Vuklia
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  • If I'm not mistaken, you never used the fact that $B$ was finite. And the indexing would entail to me that those chains are increasing. – Not Mike Mar 16 '18 at 18:35
  • I think I used the finiteness in the fact that I could simply pick the smallest of those decreasing chains, not having to worry about the infimum. Also, it was quite convenient that I could work with finite sequences. – Sha Vuklia Mar 16 '18 at 18:39
  • The entire point of the problem is to show that there exists such a chain which is finite. You seem to have assumed what you are trying to prove. – Not Mike Mar 16 '18 at 18:41
  • Every such chain is finite, because the elements in the chain are all distinct. The point is to pick an $x$ for which the length of its maximal chain is the smallest, and then show that the length is just 2. – Sha Vuklia Mar 16 '18 at 18:43
  • Consider the relation $R$ defined on the set ${a,b,c}$ by $a R b R c R a$. The set is finite but every $R$-chain is infinite. So there is something missing from your argument. – Not Mike Mar 16 '18 at 18:48
  • I'm not sure what you mean by an $R$ chain. I don't even know if what I called a "decreasing chain" is called as such. What I just meant is a sequence $(0,a_1,\dots,a_t,x)$, such that $0<a_1<\dots<a_t<x$. All those elements must be distinct, therefore what I call a decreasing chain must be finite. – Sha Vuklia Mar 16 '18 at 18:54
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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.

SphericalTriangle
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