0

I was reading this article.

And got stuck understanding this derivation:

Consider the alphabet $\Sigma = \{a,b,c\}$. A possible dependency relation is \begin{array}{ccc} D&=&\{a,b\}\times\{a,b\}\quad\cup\quad\{a,c\}\times\{a,c\}\\ &=&\{a,b\}^2\cup\{a,c\}^2\\ &=&\{(a,b),(b,a),(a,c),(c,a),(a,a),(b,b),(c,c)\}. \end{array}

Image.

How can a union of square of 2 sets derived to the 3rd line?

PS: I'm not from Maths background.

Andrés E. Caicedo
  • 79,201
kishoredbn
  • 311
  • Do you know what the square of a set is? – James Jun 08 '16 at 02:47
  • @James Yes. If it is one and the same thing mentioned here: https://en.wikipedia.org/wiki/Cartesian_product – kishoredbn Jun 08 '16 at 02:49
  • Indeed, so ${a,b}^2 = {(a,b),(b,a),(a,a),(b,b)}$ and ${a,c}^2 = {(a,c),(c,a),(a,a),(c,c)}$ and so their union is just all of these put together. – James Jun 08 '16 at 02:51
  • hey I got it. I just need to reorder them all. but Thanks anyways. – kishoredbn Jun 08 '16 at 02:51
  • "I just need to reorder them all..." technically you don't because for sets order is unimportant. The set ${1,2,3}$ is the same as the set ${3,1,2}$ and ${2,1,3}$. When order matters, we call it an ordered set. – JMoravitz Jun 08 '16 at 02:53
  • Gotcha. What you meant. Thanks a lot. – kishoredbn Jun 08 '16 at 02:54

0 Answers0