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I am reading from Halmos naive set theory, for Cartesian product, defined as: $$ A\times B=\left\{x: x \in P(P(A\cup B))\,\wedge\,\exists a \in A,\exists b \in B,\, x=(a,b)\right\} $$ Empty set belongs to $P(P(AUB))$, but since in $(a,b)$ for some $a$ in $A$ and some $b$ in $B$ , because of existence quantifier empty set doesn't belong to $A\times B$ right? But still empty set could be an ordered pair.
in Halmos, $(a,b)=\{\{a\},\{a,b\}\}$ , so empty set also an ordered pair but no cartesian product has it as an element. am i correct?

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    $\begingroup$ Any ordered pair has elements, and the empty set has not. So the empty set is not an ordered pair. Elements of Cartesian product are ordered pairs, hence are not empty. Be careful with "belong to". It gives confusion between "is element of" and "is subset of". $\endgroup$ – drhab May 27 '15 at 10:27
  • $\begingroup$ is it mentioned that ordered pair should have elements? $\endgroup$ – Arjun May 27 '15 at 10:30
  • $\begingroup$ @Arjun: "for some" $a \in A$ and "some" $b \in B$. The word "some" refers to existence. $\endgroup$ – Clive Newstead May 27 '15 at 10:31
  • $\begingroup$ This really highlights out some problems with ZFC theory. I define $A\times B$ as follows: $$\forall x,y:[(x,y)\in A\times B \iff x\in A \land y\in B]$$ $\endgroup$ – Dan Christensen May 27 '15 at 18:03
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The empty set is not an ordered pair; neither intuitively (what things would form the ordered if you just have an empty set?) nor in the formalization you recall.

Note that $(a,b)= \{\{a\}, \{a,b\}\}$ implies $\{a\} \in \{\{a\}, \{a,b\}\}$. Whence the set $(a,b)$ cannot be empty, as it has $\{a\}$ as element.

Let me add though that a Cartesian product can be equal to the empty set. Namely $ A \times B = \emptyset $ when $A= \emptyset $ or $B = \emptyset$.

Yet even in this case the empty set is not an ordered pair, as in this case there are just no ordered pairs at all.

So, $ \emptyset= A \times B $ is possible, $\emptyset \in A \times B$ is impossible.

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  • $\begingroup$ but we also take {a} itself as ordered, its not needed to take 2 distinct elements, So I thought for single element its already ordered, if no elements nothing to order. But if in ordered pair definition will they say there exists some a and some b , (a,b)={{a{,{a,b}} ? if that is the case then empty set cant be an ordered pair. did you mean this? $\endgroup$ – Arjun May 27 '15 at 10:38
  • $\begingroup$ It is true that the elements need not be different however, even for $a= b = \emptyset$ you have that $(\emptyset, \emptyset) = \{\{\emptyset\} \} $ which has as element $\{\emptyset\}$ and thus is not empty. On the other point, this is what I say in the update: if there are no elements, there is nothing to oder, yes, and the Cartesian product is empty. But this does not mean it has the empty set as element. $\endgroup$ – quid May 27 '15 at 10:42
  • $\begingroup$ i have this doubt because if {a} is an ordered pair then we are allowing B to be empty (in one sense, I am not talking about a∈A and a∈B ,so {a} is ordered ). I am saying {a} can also mean a∈A and since no b∈B, we get {a}. So i thought when both A and B are empty we get ∅, so its ordered $\endgroup$ – Arjun May 27 '15 at 10:56
  • $\begingroup$ @Arjun "then we are allowing $B$ to be empty" If $B$ is empty then $A\times B=\varnothing$ (as said by quid). So then there are no ordered pairs that belong to it. Second note: $\{a\}$ can only be an ordered pair if $a=\{c\}$ for some $c$. Then $\langle c,c\rangle=\{\{c\},\{c,c\}\}=\{\{c\}\}=\{a\}$. $\endgroup$ – drhab May 27 '15 at 11:11
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    $\begingroup$ @Arjun you can essentially say (for all a) (for all b) the set {{a}, {a,b}} is an ordered pair. However a set {{a}, {a,b}} is never the emptyset because it always has {a} as element, so for all a and for all b, {a} is an element of {{a}, {a,b}}. Therefore {{a}, {a,b}} is never the emptyset. This is not altered by the fact that you could quantifiy a over the emptyset, in which case there are no order pairs at all. But if there are none at all, again none is equal to the empty set. There is none, so how should one be equal to the empty set. $\endgroup$ – quid May 27 '15 at 12:01
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With the definition, $$A \times B = \{ X \in P(P (A \cap B)) : \exists a \in A, \exists b \in B[X = (a,b)]\}$$ One should note that we are defining with $\in$, which does not consider the empty set, but $\{\varnothing\} \in A$ is certainly possible, i.e. the set containing the empty set can be an element (also note that there are set-theoretic definition of $0$ such as $\{\varnothing\})$. However, the empty set is still a subset of $A \times B$, which differs from being an element.

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