TITLE: Independence of sigma-algebras
QUESTION [6 upvotes]: Good day to everyone. 
While solving some problem of studying character I obtained some statement to prove, which is like following (this is my internal interest to prove it rigorously).
Assume that $X_1, \ldots, X_k, X_{k+1}, \ldots, X_\ell, X_{\ell + 1}, \ldots, X_n$ are i.i.d. random variables of unknown distribution ($1 \le k < \ell < n$). See also EDIT1 for crucial assumtions.
What is really intuitively understandable, but pretty unclear how to write down, is the next statement: prove that the events
$$
\max(X_1, \ldots, X_k) \le \max(X_{k+1}, \ldots, X_\ell)
$$
and
$$
\max(X_1, \ldots, X_{\ell}) \le \max(X_{\ell+1}, \ldots, X_n)
$$
are independent.
Qualitatively and intuitively it's ok: the second one "forgets" about the configuration of the maximums in $X_1, \ldots, X_\ell$.
It's also pretty clear how to depict it "with picture": the algebra of events of the probability subspace of the first $\ell$ coordinates generates the cylindric algebra, which is subalgebra of events of the whole ($n$-coordinate) probability space.
Somehow I have the feeling that conditioning on such cylindric subalgebra should not change the distribution, but I have a lack of imagination (and knowledge of the results) now, to get something out of that.
Can anybody explain that? Thanks.
EDIT1 Due to the comments I got a clear understanding that this is not true in general, so assume that the distribution is continuous, according to the comments. Is it possible to say anything in this case?

REPLY [3 votes]: Take $X,Y,Z$ i.i.d with Bernoulli(0.5) distribution. Then $P(X \leq Y) = \frac{3}{4}$ and $P(\max(X,Y)=1) = \frac{3}{4}$ so $P(\max(X,Y) \leq Z) = \frac{5}{8}$. But
$$
P(X\leq Y, \max(X,Y) \leq Z) = P(X \leq Y, Y \leq Z) = \frac{1}{2} \neq \frac{15}{32}.
$$