Learning about subsets I have a set $A$ defined to be the power set P(X) where X is itself a nonempty set and that for the set $A$ and the relation $R$ on $A$ we have $$\forall(x,y) \in A \times A$$ $$\exists xRy$$ provided that$$ x ⊊y$$ or $$y ⊊x$$ Is the relation reflexive, symmetric and/or transitive. 
My attempt: It is not reflexive because by definition the set $x \neq y$.
It is not symmetric because one of the two statements is necessarily true and the other is then necessarily not true. ie: if we have $ x ⊊y$ then we have $\exists y_{n}\in y$ such that $y_{n} \notin x$ and then $y ⊊x$ cannot be true. I believe it is transitive because if $y ⊊x$ and $z ⊊y$ then we can just say $z ⊊x$. I'm scared because of the power set and that I may have figured this wrong... Also it's hard for me to conceptualize a relation that is transitive yet isn't reflexive or symmetric.    
Edit: I just realized $<$ fulfils this.
 A: Note that the Relation $R$ is not transitive in general. For example, take $X=\{1,2\}$, then $\mathcal{P}(X)=\{\emptyset,\{1\},\{2\},X\}$. Then we have 


*

*$\{1\}R\emptyset$ (because $\emptyset\subsetneq\{1\}$) and 

*$\emptyset R\{2\}$ (because $\emptyset\subsetneq\{2\}$) but 

*not $\{1\}R\{2\}$ (because $\{1\}\nsubseteq\{2\}$ and $\{2\}\nsubseteq\{1\}$), 


contrary to the definition of "transitive". $R$ is not even transitive if $X=\{x\}$ for some $x$ because $\emptyset R\{x\}$ and $\{x\}R\emptyset$ (because $\emptyset\subsetneq\{x\})$ would lead to $\emptyset R\emptyset$, contradicting the reflexivity. As you have shown, $R$ is never reflexive. It is symmetric though, because of the "or" in the definition of $R$. What you have shown is that the relation $\subsetneq$ is not symmetric (but transitive). But when $xRy$ means $x\subsetneq y$ or $y\subsetneq x$, then $yRx$ means $y\subsetneq x$ or $x\subsetneq y$ and we can see that these conditions are the same, hence $xRy\Longleftrightarrow yRx$.
A: Apparently it is not transitive.
You may provide a more formal proof for each part to enhance you answer but the outline is fine. 
