MHB Cartesian Product of Sets: A, B & C

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The discussion centers on the Cartesian product of sets A, B, and C, exploring whether (A × B) × C is equal to A × (B × C). It highlights that elements in (A × B) × C are ordered pairs, while those in A × (B × C) are numbers, indicating the two sets are not the same despite being isomorphic. The conversation also touches on the nature of sets being both numbers and pairs, concluding that under certain definitions, they are distinct. An example using the empty set is presented to verify the inequality of the two forms of Cartesian products. The participants confirm the correctness of the reasoning and calculations involved.
evinda
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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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