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Let $A = \{x,y,z\}$ and $B = \{1,2,3\}$.

The power set of $A$ is $$ P(A) = \{∅, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\}. $$

When working out the Cartesian product for example $A \times B$ you simply order the pairs in which the first element comes from A and the second elements come from B.

A x B = {$(x,1),(x,2),(x,3),(y,1),(y,2),(y,3),(z,1),(z,2),(z,3)$}

B x A = {$(1,x),(1,y)...(3,z)$}

For the Cartesian product $P(A)\times A$ you can not create a list of ordered pairs as $P(A)$ has a list of lists and $A$ has a list of elements. Is there a way to find the Cartesian product of $P(A)\times A$?

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  • $\begingroup$ It seems to me that the set would consist of $$\{(x, y)\mid x\in P(A) \land y \in A\}$$ That is, the ordered pair consists of (set in P(A), element in A). $\endgroup$ – Namaste Feb 8 '18 at 22:37
  • $\begingroup$ So, for example (∅,x), ({x},x), {{y},x), ({z},x), ({x,y},x). This makes sense. $\endgroup$ – Danny Sanderson Feb 8 '18 at 22:42
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The set would consist of $$P(A)\times A = \{(a, b)\mid a\in P(A) \land b \in A\}$$ That is, the ordered pair consists of (set (element) in $P(A)$, element in $A$).

$$P(A)\times A = \{(\varnothing, x), (\varnothing, y), (\varnothing, z), (\{x\}, x), \ldots (\{y\}, x), \ldots (\{x, y, z\}, z)\}$$

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