I am not quite sure about the Cartesian Product in combination with the empty set. Let's say:

$A := \{\{5\}\}$ and $B := \{\varnothing\}$.

What's the proper Cartesian Product? Is it $A\times B = \{(\{5\}, \varnothing), (\varnothing, \{5\})\}$ or simply $A\times B = \{\varnothing\}$ because of $A\times \varnothing = \varnothing$?


As Brian M. Scott, Mauro ALLEGRANZA and amWhy said:


is right. Thank you really much!

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    $\begingroup$ $A \times B = \{ ( \{ 5 \}, \emptyset) \}$ because $A \times B = \{ (x,y) : x \in A \land y \in B \}$ $\endgroup$ – Mauro ALLEGRANZA Oct 29 '14 at 19:44
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    $\begingroup$ $A \times B \ne \{ \emptyset \}$ because $B \ne \emptyset$; $B$ has one element : the emptyset, while the emptyset has no elements. $\endgroup$ – Mauro ALLEGRANZA Oct 29 '14 at 19:46

Each of the sets $A$ and $B$ has a single element. For a moment call those elements simply $a$ and $b$ to avoid being distracted by the specific nature of the elements; then it should be clear that the only member of $A\times B$ is the ordered pair $\langle a,b\rangle$. Now let’s go back and recall what $a$ and $b$ actually are: $a=\{5\}$, and $b=\varnothing$. Thus, $A\times B$ contains just the ordered pair $\langle\{5\},\varnothing\rangle$:

$$A\times B=\{\langle\{5\},\varnothing\rangle\}\;.$$

  • $\begingroup$ Isn't it A×B = {({5},∅),(∅,{5})} (two combinations)? $\endgroup$ – Conjury Oct 29 '14 at 19:46
  • $\begingroup$ @Conjury: No. By definition $A\times B$ is the set of ordered pairs $\langle a,b\rangle$ such that $a\in A$ and $b\in B$. In this case $\varnothing\notin A$, so $\langle\varnothing,\{5\}\rangle$ is not in $A\times B$. $\endgroup$ – Brian M. Scott Oct 29 '14 at 19:48
  • $\begingroup$ @Conjury: You’re welcome! $\endgroup$ – Brian M. Scott Oct 29 '14 at 19:54

$A, B$ are each a one element set. $A$ contains the set $\{5\}$ as its only element, and $B$ contains $\varnothing$ as its only element. So $$A\times B = \{(\{5\}, \varnothing)\}$$


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