1
$\begingroup$

Can I apply the cartesian product definition in the following way? For instance, in the proof below I go from "$x\in A_i \cap B_j$, for some $(i,j)\in I\times J$" to "$x\in A_i \cap B_j$, for some $i\in I$ and $j\in J$", and vice versa.

"$\bigcup_{(i,j)\in I\times J} (A_i \cap B_j) = (\bigcup_{i\in I} A_i) \cap (\bigcup_{j\in J} B_j)$"

Proof.

"$\subseteq$" Let $x \in \bigcup_{(i,j)\in I\times J} (A_i \cap B_j)$, then $x\in A_i \cap B_j$, for some $(i,j)\in I\times J$. Therefore $x\in A_i \wedge x\in B_j$, for some $i\in I$ and $j\in J$. By definition of union, $x\in \bigcup_{i\in I} A_i \wedge x\in \bigcup_{j\in J} B_j$. Then $x\in \bigcup_{i\in I} A_i \cap \bigcup_{j\in J} B_j$.

"$\supseteq$" Suppose $x\in \bigcup_{i\in I} A_i \cap \bigcup_{j\in J} B_j$, then $x\in \bigcup_{i\in I} A_i \wedge x\in \bigcup_{j\in J} B_j \Rightarrow x\in A_i \wedge x\in B_j$, for some $i\in I$ and $j\in J$. Consequently, $x\in A_i \cap B_j$, for some $i\in I$ and $j\in J$. Thus, $x\in A_i \cap B_j$, for some $(i,j)\in I\times J$. Which implies that $x\in \bigcup_{(i,j)\in I\times J} (A_i \cap B_j). \Box$

Also, if you have any suggestion of improvement, or you catch an error in the proof, please let me know. Thank you.

$\endgroup$
3
  • 1
    $\begingroup$ Your proof is fine. $\endgroup$
    – Fimpellizzeri
    Apr 18, 2017 at 3:53
  • $\begingroup$ $\bigcup_{(i,j)\in I\times J}$ is, for me, one of those things that both looks odd and is perfectly clear. Thankfully, the latter is the only important part. $\endgroup$
    – Malice Vidrine
    Apr 18, 2017 at 8:41
  • $\begingroup$ Just literally what I said. As a notation, having a compound term like $(i,j)$ in the subscript of $\bigcup$ looks strange to me. In spite of looking strange, I perfectly understand what is meant by it. Perhaps I misunderstood the nature of your question and this simply isn't relevant; feel free to ignore it in any case. $\endgroup$
    – Malice Vidrine
    Apr 18, 2017 at 9:04

1 Answer 1

Reset to default
1
$\begingroup$

Theorem:

$$\bigcup_{(i,j)\in I\times J} \left( A_i \cap B_j \right) = \left( \bigcup_{i\in I} A_i \right) \cap \left( \bigcup_{j\in J} B_j \right)$$

Proof:

$$\begin{array}{rcl} \displaystyle \bigcup_{(i,j)\in I\times J} \left( A_i \cap B_j \right) &=& \left\{ x \in A_i \cap B_j \ | \ (i,j) \in I \times J \right\} \\ &=& \left\{ x \in A_i \land x \in B_j \ | \ i \in I \land j \in J \right\} \\ &=& \left\{ x \ | \ \exists i,j[x \in A_i \land x \in B_j \land i \in I \land j \in J] \right\} \\ &=& \left\{ x \ | \ \exists i[x \in A_i \land i \in I] \right\} \cap \left\{ x \ | \ \exists j[x \in B_j \land j \in J] \right\} \\ &=& \left\{ x \in A_i \ | \ i \in I \right\} \cap \left\{ x \in B_j \ | \ j \in J \right\} \\ &=& \displaystyle \left( \bigcup_{i\in I} A_i \right) \cap \left( \bigcup_{j\in J} B_j \right) \end{array}$$

$\endgroup$
1
  • $\begingroup$ @JuliusD.Galt I just looked up the definition of each operation, including union, cartesian product, and intersection... $\endgroup$
    – DHMO
    Apr 18, 2017 at 6:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .