# Cartesian product

• Aug 27th 2009, 02:50 AM
The Lama
Cartesian product
My teatcher wasn't able to explain this for me in a clear way. See if you guys can help.
Let A and B be two random sets. Prove that A $\times" alt="\times" /> B (B $\cap" alt="\cap" /> C) = (A $\times" alt="\times" /> B) $\cap" alt="\cap" /> (A $\cap" alt="\cap" /> C) by putting in the element (x,y)
• Aug 27th 2009, 04:31 AM
Defunkt
Sort of difficult to see what you want to prove like this...

Did you mean: $A \times (B \times C) = (A \times B) \times (A \times C)$ ?
• Aug 27th 2009, 05:09 AM
The Lama
Sorry about that. Didn't know how to handle the LaTex-codes right. But I think know what's going on now.

But yes, that was what I meant. The book wants me to prove the equivalence by putting in the elemet (x,y) into that.
• Aug 27th 2009, 06:32 AM
Plato
LaTex Help:
$$A \times \left( {B \times C} \right) \ne \left( {A \times B} \right) \times \left( {A \times C} \right)$$ gives $A \times \left( {B \times C} \right) \ne \left( {A \times B} \right) \times \left( {A \times C} \right)$.

Why? $A \times \left( {B \times C} \right)$ is a set of triples.
Whereas, $\left( {A \times B} \right) \times \left( {A \times C} \right)$ is a set of pairs made up of pairs.
• Aug 27th 2009, 07:17 AM
The Lama
Ok, this is how the question looks like and what the key says.

$A \times (B \cap C) = (A \times B) \cap (A \times C)$

Let (x,y) be a random element in $A \times (B \cap C)$
Then we have that X $\epsilon$ A and y $\epsilon$ B $\cap$C which give y $\epsilon$ B and y $\epsilon$C.

(x,y) $\epsilon$ A x B and (x,y) $\epsilon$ A x C so (x,y) $\epsilon$ (A x B) $\cap$ (A x C)
and now we can prove that (A x B) $\cap$ (A x C) $\subseteq$ A x (B $\cap$ C).

But I don't get a grip over this. For example why do you have to put the x in A and the y in (B $\cap$C) and not the other way around?

I would like to see this in a graphic 3D model so I could see what I was doing.
• Aug 27th 2009, 07:28 AM
Defunkt
Quote:

Originally Posted by The Lama
Ok, this is how the question looks like and what the key says.

$A \times (B \cap C) = (A \times B) \cap (A \times C)$

Let (x,y) be a random element in $A \times (B \cap C)$
Then we have that X $\epsilon$ A and y $\epsilon$ B $\cap$C which give y $\epsilon$ B and y $\epsilon$C.

(x,y) $\epsilon$ A x B and (x,y) $\epsilon$ A x C so (x,y) $\epsilon$ (A x B) $\cap$ (A x C)
and now we can prove that (A x B) $\cap$ (A x C) $\subseteq$ A x (B $\cap$ C).

But I don't get a grip over this. For example why do you have to put the x in A and the y in (B $\cap$C) and not the other way around?

I would like to see this in a graphic 3D model so I could see what I was doing.

You need to show $x \in A$ and $y \in {B \cap C}$ because that's simply the order of the cartesian product -- for example:

$X \times Y = \left\{ (x,y) : x \in X ; y \in Y\right\}$ While
$Y \times X = \left\{ (y,x) : y \in Y; x \in X\right\}$

And those are ordered pairs, so:
$(x,y) \neq (y,x) \Rightarrow X \times Y \neq Y \times X$

Also, your wording is a bit off... You should say "Let $(x,y) \in {A \times B}$", or "Let $(x,y)$ be some element in $A \times B$"; random is not quite the right word here!
• Aug 27th 2009, 07:29 AM
Plato
Quote:

Originally Posted by The Lama
(x,y) $\epsilon$ A x B and (x,y) $\epsilon$ A x C so (x,y) $\epsilon$ (A x B) $\cap$ (A x C)
and now we can prove that (A x B) $\cap$ (A x C) $\subseteq$ A x (B $\cap$ C).
But I don't get a grip over this. For example why do you have to put the x in A and the y in (B $\cap$C) and not the other way around?

More LaTex help.
$$(x,y)\in A\times (B\cap C)$$ gives $(x,y)\in A\times (B\cap C)$.

The is simply the way a cross product is defined.

$(x,y)\in W\times Z$ means $x\in W\text{ and }y\in Z$
• Aug 27th 2009, 08:51 AM
The Lama
Quote:

Originally Posted by Defunkt
Also, your wording is a bit off... You should say "Let $(x,y) \in {A \times B}$", or "Let $(x,y)$ be some element in $A \times B$"; random is not quite the right word here!

Alright. Thanks. I just tried to translate it as best I could from a swedish compendium.

yea. x and $\times$. Got a bit lazy there I guess.