Relations: is the composition of an empty set with a nonempty set empty? And the other way around Given $T\circ S=\emptyset$ and $R$ nonempty, would
$$(T \circ S) \circ R$$
be anything other than the empty set? 
I'm also curious the other way around. I think that it would be just empty. 
 A: Recall the definition of composition:
Let $R,S$ be binary relations. $R\circ S = \{\langle x,y\rangle\mid \exists z(\langle x,z\rangle\in S\wedge \langle z,y\rangle\in R)\}$. (Essentially this a generalization of composition of functions.)
Suppose $R=\emptyset$ (which is a relation as all its members are ordered pairs) then clearly $R\circ S=\emptyset$, otherwise $x(R\circ S)y$ implies there is some $z$ such that $\langle x,z\rangle\in S=\emptyset$ which is a contradiction.
A: Asaf already gave a good answer. I prefer the order of composition of relations to be the same as for functions (which, if I am not mistaking, is the opposite of Asaf's order). 
So suppose $X,Y,Z$ are sets and we have relations $X\xrightarrow{R}Y\xrightarrow{S}Z$ (by which I mean $R$ is a relation from $X$ to $Y$, and $S$ is a relation from $Y$ to $Z$). Then the composition $S\circ R$ is a relation from $X$ to $Z$ given by
$S\circ R:=\{(x,z)|\exists y\in Y: (x,y)\in R\text{ and }(y,z)\in S\}$.
Now to your question about the converse: this is not true. An easy counter-example might be to take $X=\{1\}=Z$ and $Y=\{1,2\}$, and $R=\{(1,1)\}$ and $S=\{(2,1)\}$. Then $(1,1)\notin S\circ R$. So $S\circ R=\emptyset$ while $R\neq \emptyset$ and $S\neq \emptyset$.
