0

I am currently studying for my first exam as a maths student, and is stuck on the problem below. I understood the solution when I saw it, but I am not quite sure of how to explain it properly. If anyone might be able to explain the solution, feel free to do so!

Problem

It is possible to represent a relation $R$ from $A$ to $B$ as a function $f_{R}(a) : A \rightarrow P(B),$ where $P(B)$ is the power set of $B$.

Solution

Let $f_{R}(a) = \{b \in B : aRb\} \in P(B)$ for $a \in A$.

You then have that $aRb \iff b \in f_{R}(a)$ which is a suitable representation of the relation.

statgura
  • 1
  • Hello, and welcome! I think a good way to summarise what is going on here and why it's important is "there is a natural one-to-one correspondence between relations on $A$ and $B$ and functions from $A$ to $\mathcal P(B)$, given by $R \mapsto f_R$" (you could prove it's indeed bijective if you want to "explain" it more). Have you showed us exactly what the text of the problem and the solution were? It looks a bit strange - ie the problem as stated isn't really a problem. Are you maybe just asking about something you find confusing? – Izaak van Dongen Oct 11 '23 at 15:18
  • Does this answer your question? Bijection between $P(E)^F$ and $P(E*F)$ (more specifically this) – Anne Bauval Oct 11 '23 at 15:41
  • Or more generally this, applied to $C={0,1}.$ – Anne Bauval Oct 11 '23 at 15:48

1 Answers1

0

I'm not sure what you mean by an "explanation".

The idea here is that you know the relation when for each particular $a \in A$ you know all the elements of $B$ that $a$ is related to. That's the subset of $B$ you are calling $f_R(a)$.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199