MHB Proving Reflexive, Symmetric and Transitive Properties of Relation R on P(U)

[leigh ramona]

Let U be a universal set, and let C be any subset of U. Let R be the relation on P(U) defined by A R B if $A \cap C = B \cap C$. Determine whether the relation is reflexive, symmetric, and/or transitive. Prove you answer.

 

[HallsofIvy]

First, do you know what "reflexive", "symmetric", and "transitive" mean? Write down the definitions and show that this relation satisfies those definitions.

 

[leigh ramona]

So reflexive is equal to each other. Like x R x.
Symmetric is x R y = y R x
Transitive is if x R y and y R z, then x R z.

The relation is symmetric because if A \cap C = B \cap C, then C \cap A = C \cap B. Is this correct?

The relation is also transitive, because if A \cap C and B \cap C, then A \cap B.

I'm not sure about the reflexive.

 

[Evgeny.Makarov]

So reflexive is equal to each other. Like x R x.

Your sentence does not make sense because it lacks the subject: what is equal to each other?. And when you say $xRx$, I'll ask you: what is $x$? Do you mean $xRx$ holds for some unspecified $x$, for all $x$, for some specific $x$? What set does $x$ range over?

Symmetric is x R y = y R x

This sentence is also problematic. For each $x$ and $y$, $xRy$ is either true or false. What do you mean by $xRx=yRx$? For which $x$ and $y$?

Transitive is if x R y and y R z, then x R z.

This would be correct if you added "for all $x$, $y$ and $z$".

The relation is symmetric because if A \cap C = B \cap C, then C \cap A = C \cap B. Is this correct?

What you wrote is true, but what does this have to do with $R$? Please enclose formulas in dollar signs: \$A\cap C\$.

The relation is also transitive, because if A \cap C and B \cap C, then A \cap B.

This does not makes sense because $A\cap C$ cannot be true or false: it's a set. Therefore, you can't write "If $A\cap C$...".

If you increase the level of your precision, it will help you not only to communicate more clearly, but to understand the problem and definitions better as well.