Finding whether a relation is transitive, reflexive and symmetric.

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.

• The word "group" means something specific in mathematics. Feb 6, 2021 at 18:44
• @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. Feb 6, 2021 at 18:50

$$M$$ 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.