Disjoint cartesian products

**TheProphet** · September 8th 2011, 03:12 AM

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 $C_2(x) = (5,6)$ ?

**Opalg** · September 8th 2011, 03:38 AM

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.

**TheProphet** · September 8th 2011, 04:07 AM

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.

**Opalg** · September 8th 2011, 12:15 PM

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.

**TheProphet** · September 8th 2011, 12:22 PM

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.

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