Cartesian product

**The Lama** · August 27th 2009, 02:50 AM

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 $ B (B $ C) = (A $ B) $ (A $ C) by putting in the element (x,y)

**Defunkt** · August 27th 2009, 04:31 AM

Sort of difficult to see what you want to prove like this...

Did you mean: $A \times (B \times C) = (A \times B) \times (A \times C)$ ?

**The Lama** · August 27th 2009, 05:09 AM

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.

**Plato** · August 27th 2009, 06:32 AM

LaTex Help:
[tex]A \times \left( {B \times C} \right) \ne \left( {A \times B} \right) \times \left( {A \times C} \right)[/tex] 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.

**The Lama** · August 27th 2009, 07:17 AM

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.

**Defunkt** · August 27th 2009, 07:28 AM

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!

**Plato** · August 27th 2009, 07:29 AM

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.
[tex](x,y)\in A\times (B\cap C)[/tex] 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 $x\in W\text{ and }y\in Z$

**The Lama** · August 27th 2009, 08:51 AM

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.

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