Let's say we have set: $A = \{1,2,3,...,10\}$, and there is a relation $M=P(A)-{\phi}=\{X | X\subseteq A, x\neq \phi\}.$

And there is this relation:

  • $T$ is defined as: $\{X,Y\}\in T$ iff $\{X,Y\}$ is a partition of $A$.

So to find out whether it's transitive, I first tried to find the result of the relation $M$, and then found it to be almost impossible and impractical, as calculating $2^{10}$ of subsets is not the best idea, and the by using this I could calculate the $R^2$, but that's not the case this time.

Then I tried finding whether it's Symmetric, but because I couldn't calculate $M$, as I think I can't also find $M^{-1}$.

And.. I am pretty sure it's not reflexive as it can't have the Identity because it's the $P(A)$ of this set, meaning there isn't a set like this one: $\{1,1\}$.

I would like to get some assistance \ explanation on finding whether a relation is reflexive, symmetric, and transitive.

  • $\begingroup$ The word "group" means something specific in mathematics. $\endgroup$
    – Shaun
    Feb 6, 2021 at 18:44
  • $\begingroup$ @Shaun oh sorry, English is not my native learning language, and we're naming it as "Group" and not "set" in my native learning language. $\endgroup$ Feb 6, 2021 at 18:50

1 Answer 1


$M$ isn't a relation, it's the universe on which $T$ is a relation.

Since $\{X,\,Y\}$ is only a partition of $A$ only if $X\cap Y=\emptyset$, in general $(X,\,X)\notin T$, so $T$ isn't reflexive.

Since $\{X,\,Y\}=\{Y,\,X\}$, $T$ is symmetric.

If $(X,\,Y)\in T$ and $(Y,\,Z)\in T$ then $X=A\setminus Y=Z$, so $(X,\,Z)=(X,\,X)\notin T$ (becaise $\emptyset\cup\emptyset\ne A$), so $T$ isn't transitive.

  • $\begingroup$ Oh thanks! Now that makes sense, I have been breaking my brain for the last few hours and found no way of how to answer it. $\endgroup$ Feb 6, 2021 at 19:58

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