MHB What is the cartesian product of two sets?

[evinda]

Hey! (Nerd)

If $A,B$ are sets, the unique set $\{ <a,b>: a \in A \wedge b \in B \}$ is called cartesian product of $A,B$ and is symbolized as $A \times B$.

I want to find the cartesian product $\mathbb{Z} \times \{ 1, 2 \}$.
I thought, that it is equal to $\{ <x,1>,<x,2>: x \in \mathbb{Z}\}$.

According to my notes, it is equal to: $\{ <x,1>: x \in \mathbb{Z}\} \cup \{ <x,2>: x \in \mathbb{Z}\}$.

Are these the same? (Thinking)

Also, how can we show that $\varnothing \times A=A \times \varnothing=\varnothing$, right? (Worried)

It is: $\varnothing \times A=\{ <b,a>: b \in \varnothing, a \in A \}$?
Can we conclude from this, that $\varnothing \times A=\varnothing$ ? (Thinking)

 

[I like Serena]

Hey! (Nerd)

If $A,B$ are sets, the unique set $\{ <a,b>: a \in A \wedge b \in B \}$ is called cartesian product of $A,B$ and is symbolized as $A \times B$.

I want to find the cartesian product $\mathbb{Z} \times \{ 1, 2 \}$.
I thought, that it is equal to $\{ <x,1>,<x,2>: x \in \mathbb{Z}\}$.

According to my notes, it is equal to: $\{ <x,1>: x \in \mathbb{Z}\} \cup \{ <x,2>: x \in \mathbb{Z}\}$.

Are these the same? (Thinking)

Hi! (Blush)

I have just looked it up in Set-builder notation on wiki.

It appears that formally the expression to the left of the ':' should represent one element and not two.
Since your notation is not ambiguous, I would still consider them the same.
But formally your first form is not defined. (Nerd)

Also, how can we show that $\varnothing \times A=A \times \varnothing=\varnothing$, right? (Worried)

It is: $\varnothing \times A=\{ <b,a>: b \in \varnothing, a \in A \}$?
Can we conclude from this, that $\varnothing \times A=\varnothing$ ? (Thinking)

Yes.
There is no element <b,a> that satisfies the conditions.
So the result is an empty set. (Nod)

 

[evinda]

It appears that formally the expression to the left of the ':' should represent one element and not two.
Since your notation is not ambiguous, I would still consider them the same.
But formally your first form is not defined. (Nerd)

So, is this: $\{ <a,b>,<c,d> : a,b,c,d \in \mathbb{Z}\}$ well-defined, but this: $\{ <a,b>,<a,d> : a,b,d \in \mathbb{Z}\}$ isn't? (Thinking)
Or have I understood it wrong? (Thinking)
And what if $x$ had a specific value?

Yes.
There is no element <b,a> that satisfies the conditions.
So the result is an empty set. (Nod)

So, can we say that since there is no element $b$, such that $b \in \varnothing$, there is no $<c,d>$, such that $<c,d> \in \{ <b,a>: b \in \varnothing, a \in A\}$, so $\forall <c,d>: <c,d> \notin \varnothing \times A$.
Therefore, $ \varnothing \times A=\varnothing$.

(Thinking)

 

[I like Serena]

So, is this: $\{ <a,b>,<c,d> : a,b,c,d \in \mathbb{Z}\}$ well-defined, but this: $\{ <a,b>,<a,d> : a,b,d \in \mathbb{Z}\}$ isn't? (Thinking)
Or have I understood it wrong? (Thinking)

Formally, neither is defined.
Informally, both are fine. (Smile)

And what if $x$ had a specific value?

Which x?

So, can we say that since there is no element $b$, such that $b \in \varnothing$, there is no $<c,d>$, such that $<c,d> \in \{ <b,a>: b \in \varnothing, a \in A\}$, so $\forall <c,d>: <c,d> \notin \varnothing \times A$.
Therefore, $ \varnothing \times A=\varnothing$.

(Thinking)

Yes, we can say that. (Mmm)

 

[evinda]

Formally, neither is defined.

Could you explain me further, why formally, neither is defined? (Thinking)

Which x?

From $\{ <x,1>, <x,2>: x \in \mathbb{Z} \}$..