0

Every function $f: X \rightarrow Y$ is actually a relation between $X$ and $Y$: $$R_f = \{ (x, f(x)) : x \in X \}$$

If $X$ and $Y$ are fields, we can take sums and products of two functions and say $$R_f + R_g = R_{f + g} = \{ (x, f(x) + g(x)) : x \in X\}$$ $$R_f * R_g = \{ (x, f(x) g(x)) : x \in X\}$$

But not every relation is a function, so it would be nice to add and multiply relations that are not functions. I suspect we can define the sum of two arbitrary relations between X and Y by $$R + S = \{ (x, y_1 + y_2): (x, y_1) \in R, (x, y_2) \in S \}$$ and the product by $$R S = \{ (x, y_1 y_2): (x, y_1) \in R, (x, y_2) \in S \}.$$

Is this a good/correct definition? I assume this has been done before somewhere but I can't find it.

Paul Castle
  • 761
  • can you explain what is $R$ and $S$ as subset and thinks – m.idaya Jun 24 '16 at 14:17
  • I'm taking R and S to be arbitrary relations, so each one could be any subset of X x Y. – Paul Castle Jun 24 '16 at 14:21
  • in this case , the definition of R+S and RS not complet, we must thinks at $(x_1,y_2)\in R$ and $(x_2,y_2)\in S $ – m.idaya Jun 24 '16 at 14:25
  • I don't think that works, if I understand you correctly. Or at least, it doesn't satisfy $R_f + R_g = R_{f + g}$. – Paul Castle Jun 24 '16 at 14:33
  • if you have laws on $X$ and on $Y$, I think it's possible to induced such laws in the set of relations from $X$ to $Y$ ( that is a subset of $X\times Y$) – m.idaya Jun 24 '16 at 14:43
  • he is not thinking about the intersection and to the meeting since it is a set problem and not algebraic problem, and then speaking about compatibility if in additional structure are of X and Y – m.idaya Jun 24 '16 at 14:51

0 Answers0