Let $X$ be the Cantor space $2^\omega$ and $Clop(X)$ be the field of open-closed subsets of $X$. It is known that $Clop(X)$ is incomplete, and a minimal completion of $Clop(X)$ is e.g. the regular open algebra $RO(X)$ of $X$.
$Clop(X)$ embeds naturally into $RO(X)$ because every open-closed set is regular open. So what really matters for the completion is the case involving infinite operations. But what is the point of even considering such infinite operations if they are not defined in $Clop(X)$ in the first place?
It is a question that eludes me. Any input would be appreciated. Thanks.
One interesting application of the notion of a complete Boolean algebra is this: measure algebras are complete. Hmm, you wanted to know what good is completeness; this is just an interesting instance of completeness. Look at it this way:: What's below is an interesting and not at all trivial fact about measure theory, and the proof is an application of the completeness of the lattice of orthogonal projections on a Hilbert space.
Or: The interesting fact about measure theory is an application of, in fact is equivalent to, the completeness of the algebra of clopen subsets of the maximal ideal space of $L^\infty$.
Anyway, ($\sigma$-finite) measure spaces are complete. In other words: Say $(X,\Sigma,\mu)$ is a $\sigma$-finite measure space; in particular $\Sigma$ is a $\sigma$-algebra on $X$. Now $\Sigma$ is not closed under arbitrary unions, but an arbitrary subset of $\Sigma$ does have a least upper bound modulo null sets, in this sense:
That is, $E$ almost contains every $E_\alpha$, and any other measurable set that almost contains every $E_\alpha$ almost contains $E$; so in a sense $E$ is almost the union of the $E_\alpha$.
Sketch of proof if $\mu(X)<\infty$ (the $\sigma$-finite case follows easily):
Say $V_\alpha$ is the subspace of $L^2(\mu)$ consisting of the functions that vanish almost everywhere on $X\setminus E_\alpha$. Let $V$ be the closed span of the $V_\alpha$ in $L^2$.
It's easy to verify that if $f\in V$ and $\phi\in L^\infty$ then $\phi f\in V$. It's not hard to show from this that if $P:L^2\to V$ is the orthogonal projection then $P(\phi f)=\phi P(f)$ for $f\in L^2$ and $\phi\in L^\infty$.
Now $L^2\subset L^1$, so $f\mapsto\int Pf$ defines a bounded linear functional on $L^2$; hence there exists $m\in L^2$ such that $$\int Pf=\int mf\quad(f\in L^2).$$If $\phi\in L^\infty$ and $f\in L^2$ then $\int\phi Pf=\int P(\phi f)=\int\phi mf;$ since $Pf,mf\in L^1$ this shows that $$Pf=mf\quad(f\in L^2).$$(Not that we need to say this at this point, but in particular $||mf||_2\le||f||_2$ for all $f\in L^2$, hence $||m||_\infty\le 1$.) Since $P^2=P$ we must have $m^2=m$, so there exists $E$ with $m=\chi_E$; the result follows. (Show first that $V$ is the space of all $f\in L^2$ which vanish almost everywhere on $X\setminus E$.)
Edit: Actually I was bluffing somewhat there - recalled that you could do this this way but didn't quite recall the punchline. The final "the result follows" goes like so:
First, since $P$ is the orthogonal projection onto $V$ and $V_\alpha\subset V$, we have $Pf=f$ for every $f\in V_\alpha$. In particular $\chi_E\chi_{E_\alpha}=\chi_{E_\alpha}$, so $\mu(E_\alpha\setminus E)=0$.
Now suppose that $\mu(E_\alpha\setminus E')=0$ for every $\alpha$. Then $\chi_{E'}f=f$ for every $f\in V_\alpha$, and hence for every $f\in V$. But $P\chi_E=\chi_E^2=\chi_E$, hence $\chi_E\in V$. So $\chi_{E'}\chi_E=\chi_E$, hence $\mu(E\setminus E')=0$.