1. ## Disjoint cartesian products

Let $(E,\mathcal{E})$ and $(F,\mathcal{F})$ be two measurable spaces.

Let $\mathcal{E} \otimes \mathcal{F} = \sigma(\mathcal{E} \times \mathcal{F})$, where $\mathcal{E} \times \mathcal{F} = \{A \subset E \times F : A = \Lambda \times \Gamma, \Lambda \in \mathcal{E}, \Gamma \in \mathcal{F}\}$.

For $C \in \mathcal{E} \otimes \mathcal{F}$, let $C(x) = \{y : (x,y) \in C \}$. If we now let $C_{n}$ be a pairwise disjoint sequence in $\mathcal{E} \otimes \mathcal{F}$, then the $C_{n}(x)$ are also pairwise disjoint. <- HERE IS MY QUESTION.

How can this be true? For example, if $C_{1}(x) = (1,2) \times (5,6)$ and $C_{2}(x) = (3,4) \times (5,6)$, then $C_{1}$ and $C_2$ are disjoint, because no ordered pair $(a,b)$ is in both $C_1$ and $C_2$. But, $C_{1}(x) = (5,6)$ and $C_2(x) = (5,6)$ ?

2. ## Re: Disjoint cartesian products

Originally Posted by TheProphet
Let $(E,\mathcal{E})$ and $(F,\mathcal{F})$ be two measurable spaces.

Let $\mathcal{E} \otimes \mathcal{F} = \sigma(\mathcal{E} \times \mathcal{F})$, where $\mathcal{E} \times \mathcal{F} = \{A \subset E \times F : A = \Lambda \times \Gamma, \Lambda \in \mathcal{E}, \Gamma \in \mathcal{F}\}$.

For $C \in \mathcal{E} \otimes \mathcal{F}$, let $C(x) = \{y : (x,y) \in C \}$. If we now let $C_{n}$ be a pairwise disjoint sequence in $\mathcal{E} \otimes \mathcal{F}$, then the $C_{n}(x)$ are also pairwise disjoint. <- HERE IS MY QUESTION.

How can this be true? For example, if $C_{1}(x) = (1,2) \times (5,6)$ and $C_{2}(x) = (3,4) \times (5,6)$, then $C_{1}$ and $C_2$ are disjoint, because no ordered pair $(a,b)$ is in both $C_1$ and $C_2$. But, $C_{1}(x) = (5,6)$ and $C_2(x) = (5,6)$ ?
Your example is obviously correct. Given disjoint sets in a product space, it certainly does not follow that their projections onto one of the coordinate spaces are disjoint.

3. ## Re: Disjoint cartesian products

Ok. The reason why I asked is because they make this claim in a proof in a book I´m reading. This is a big error, or it is something else that I am missing.
Thank you anyway.

4. ## Re: Disjoint cartesian products

Oops, I was wrong first time, and the book is right. The fact that x appears in the notation C(x) implies that x is meant to be fixed. So the sets $C_n(x)$ are the slices of the sets $C_n$ obtained by fixing the first coordinate x and letting the y coordinate vary.

Originally Posted by TheProphet
For example, if $C_{1}(x) = (1,2) \times (5,6)$ and $C_{2}(x) = (3,4) \times (5,6)$, then $C_{1}$ and $C_2$ are disjoint, because no ordered pair $(a,b)$ is in both $C_1$ and $C_2$. But, $C_{1}(x) = (5,6)$ and $C_2(x) = (5,6)$ ?
In that example, you should have taken the sets to be $C_{1} = (1,2) \times (5,6)$ and $C_{2} = (3,4) \times (5,6)$ (in other words, the parameter x does not appear in the names of the product sets). If you now choose x = 1.5, for example, then $C_1(1.5) = (5,6)$, but the set $C_2(1.5)$ is empty, because $C_2$ does not contain any points with x-coordinate 1.5.

5. ## Re: Disjoint cartesian products

Ah, thank you. I actually meant $C_1 = (1,2) \times (5,6)$ and $C_2 = (3,4) \times (5,6)$, this was a misprint on my part. Sorry about that.