MHB Cartesian Product of Sets: A, B & C

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Hi! (Wave)

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

If $A,B,C$ sets, then we define the Cartesian product of $A,B,C$ as:

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

But.. is it: $(A \times B) \times C=A \times (B \times C)$, or not? (Thinking)
 
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Suppose that $A=B=C=\Bbb N$. If $x\in (A\times B)\times C$, then the first component of $x$ is an ordered pair. If $x\in A\times (B\times C)$, then the first component of $x$ is a number. And yes, in set theory both ordered pairs and numbers are sets and it may happen (or not?) that a set is both a number and a pair. But it should be easy to find a number that is not a pair and vice versa.
 
Evgeny.Makarov said:
Suppose that $A=B=C=\Bbb N$. If $x\in (A\times B)\times C$, then the first component of $x$ is an ordered pair. If $x\in A\times (B\times C)$, then the first component of $x$ is a number. And yes, in set theory both ordered pairs and numbers are sets and it may happen (or not?) that a set is both a number and a pair. But it should be easy to find a number that is not a pair and vice versa.

Can it only happen that a set is both a number and a pair, if the pair contains twice the same number? Or am I wrong? (Thinking)
 
Whether a set can be both an ordered pair and a number depends on the definitions of pairs and numbers. If we are talking about Kuratowski definition of pairs: $(a,b)=\{\{a\},\{a,b\}\}$, and Von Neumann definition of ordinals (numbers), then consider $(\varnothing,\varnothing)=\{\{\varnothing\}\}$. This set is not a Von Neumann ordinal. In fact, the only Von Neumann ordinal with one or two elements are $2=\{\varnothing\}$ and $2=\{\varnothing,\{\varnothing\}\}$. They are different from an ordered pair $p=\{\{a\},\{a,b\}\}$ because $\varnothing\in1$ and $\varnothing\in2$, but $\varnothing\notin p$. So for these definitions, a number is never an ordered pair.

Even if it were possible for a set to be both a number and a pair, that would be an incident of encoding of pairs and numbers. I wrote in the thread about Kuratowski pairs that it is merely a hack. Conceptually, an ordered pair is a completely different object from a natural number. And since elements of $(A\times B)\times C$ have pairs as their first component and elements of $A\times (B\times C)$ have, say, numbers as their first component, these sets are different. They are isomorphic, though.
 
$$A^3=(A \times A) \times A$$

When $w \in A^3$, to see of which form it is, do we have to do it like that?

It will be of the form $<x,y>$, where $x \in A \times A$ and $y \in A$.
Since, $x \in A \times A$, it is of the form $<c,d>: c,d \in A$.

Therefore, $w=<<c,d>,y>:c,d,y \in A $.

Or am I wrong? (Thinking)
 
You are correct.
 
Evgeny.Makarov said:
You are correct.

Nice, thank you very much! (Smile)
 
Evgeny.Makarov said:
Suppose that $A=B=C=\Bbb N$. If $x\in (A\times B)\times C$, then the first component of $x$ is an ordered pair. If $x\in A\times (B\times C)$, then the first component of $x$ is a number. And yes, in set theory both ordered pairs and numbers are sets and it may happen (or not?) that a set is both a number and a pair. But it should be easy to find a number that is not a pair and vice versa.

I want to verify, that $X \times (Y \times Z) \neq (X \times Y) \times Z$, for $X=\{ \varnothing \},Y=\{ \varnothing \}, Z=\{ \varnothing, \{ \varnothing \} \}$.

Is it like that?

$$X \times (Y \times Z)=\{ \{ \varnothing \} \times (\{ \varnothing \} \times \{ \varnothing,\{ \varnothing \} \}) \}=\{ \{ \varnothing \} \times (<\varnothing, \varnothing>,<\varnothing,\{ \varnothing \}>) \}=\{ <\varnothing,<\varnothing, \varnothing>>, <\varnothing,<\varnothing,\{ \varnothing \}>>\}$$

$$(X \times Y) \times Z=\{ (\{ \varnothing \} \times \{ \varnothing \}) \times \{ \varnothing,\{ \varnothing \} \} \}=\{ <\{ \varnothing \}, \{ \varnothing \}> \times \{ \varnothing, \{ \varnothing \} \}\}=\\ =\{ << \{ \varnothing\},\{ \varnothing\}>, \varnothing\},<< \{ \varnothing \}, \{ \varnothing \}>,\{ \varnothing \}> \}$$Or have I done something wrong? (Thinking)
 

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