Showing that any finite algebra is generated by a collection of disjoint sets

Question

As the title says, I am trying to prove that any finite algebra $\mathcal A$ is generated by a collection of disjoint sets.

I am not quite sure where to start with this proof. I am familiar with generating algebras, so I started of by saying that if an algebra $\mathcal A \subseteq \mathcal P(X)$ is generated by a collection of sets $\mathcal E$, then $\mathcal A$ is the intersection of all algebras, which contain $\mathcal E$. Now I want to show that the elements of $\mathcal E$ must be disjoint, or otherwise $\mathcal A$ would not be an Algebra, but how do I do this?

EDIT:

I changed my approach and ended up with a proof like this:

Let $\mathcal A$ be an arbitrary finite algebra. I want to show that there exists a collection of disjoint sets $\mathcal E$, which generates $\mathcal A$. I start off with choosing $\mathcal E = \mathcal A$. Clearly, $\mathcal E$ generates $\mathcal A$. Since $ X$ and $\emptyset \in \mathcal A$ for any algebra, I can remove these from $\mathcal E$. Furthermore, since the union of finitely many sets is contained in an algebra, that contains those finitely many sets, I can also remove all of these finitely many sets from $\mathcal E$. Therefore I end up with a collection of set $\mathcal E$, which is disjoint.

Is this correct, and if so: How can I make this more formal?

Answer 1

Let $\mathcal{A}$ be an algebra on $X$ and let $E_x$ be the intersection of all elements of $\mathcal{A}$ that contains $x$ (since $\mathcal{A}$ is finite, this intersection is finite and then $E_x$ belongs to the algebra $\mathcal{A}$). Now you can look at those $E_x$'s that are distinct and make a family $\{E_x\}$ that forms a partition of $X$ by elements of $\mathcal{A}$ (try to prove this, I can give more hints if needed). Finally, if $F\in\mathcal{A}$, you can see that $$F=\bigcup_{x\in F} E_x.$$ So $\mathcal{A}$ is generated by $\{E_x\}$.

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I will give one proof that the family $\{E_x\}$ is disjoint. Maybe you can find another.

Suppose that there exists $z\in E_x\cap E_y$. Then $z\in E_x$, what implies $E_z\subset E_x$. If $x\notin E_z$, then $\emptyset \neq E_x\backslash E_z\in \mathcal{A}$ and $x\in E_x\backslash E_z \subsetneq E_x$, what is an absurd because $E_x$ is the smallest element of $\mathcal{A}$ that contains $x$. So $x\in E_z$, what implies $E_x\subset E_z$, that is, $E_x=E_z$. In the same way you can proof that $E_y=E_z$. Therefore $E_x=E_y$ and we're done.

Answer 2

It looks like you're misunderstanding the task. Your goal is written as

Any finite algebra $\mathcal A$ is generated by a collection $\mathcal E$ of disjoint sets.

and your ansatz looks like you're interpreting that as

Whenever we have a collection $\mathcal E$ of sets that generates some finite algebra $\mathcal A$, then the sets in $\mathcal E$ are disjoint.

This is not what is asked for, though -- and it is not even true. Consider for example that $\mathcal E = \{\{0\},\{0,1\}\} \subseteq \mathcal P(\{0,1,2\})$ generates $\mathcal P(\{0,1,2\})$ itself but is not disjoint.

What you're actually supposed to prove is

Whenever we have a finite algebra $\mathcal A$ then there exists a collection $\mathcal E$ of disjoint sets such that $\mathcal E$ generates $\mathcal A$.

So the first part of the answer should be to find a way, given only $\mathcal A$ itself, to make an $\mathcal E$ that will work. It is not given to you in advance.

(Your underlying confusion may be that you assume that every algebra has exactly one collection that generates it. That is not the case; the same algebra can be generated in many different ways).