Doubling checking my understanding of these set relations Just trying to make sure that my understanding of the set relations below is correct  
T1 = {x $\in$ $\Delta$ | for every y, (x,y) $\in$ R implies y $\in$ C}
T2 = {x $\in$ $\Delta$ | there exists y, (x,y) $\in$ R and y $\in$ C}
$\Delta$ = {a, b, c, d, e, f, g}
R = {(a,b), (c,b), (d,e)}
C = {b, e}
T1 = {e}
T2 = {a, c, e}
Or is this wrong?
 A: Well $x\in T_1$ iff for all $y\in\Delta$ either $(x,y)\in R$ and $y\in C$ or $(x,y)\notin R$. In other words $x\notin T_1$ iff there exists $y\in Y$ such that $(x,y)\in R$ and $y\notin C$. But because for every $(x,y)\in R$, $y\in C$ this shows that $T_1=\Delta$.
Now for the second set, it is easily seen to be $\{a,c,d\}$.
A: Rewrite $T_1$ as:
$$T_1=\Delta-\left\{x\in\Delta\mid\exists y,(x,y)\in R\land y\notin C\right\}$$
It is easy to see that to satisfy the first condition, $y$ has to be either $b$ or $e$, but both are in $C$ and thus violate the second condition. Therefore there is no such set and we can write:
$$T_1=\Delta-\emptyset=\Delta$$
Rewrite $T_2$ as:
$$T_2=\Delta-\left\{x\in\Delta\mid\forall y,(x,y)\notin R\lor y\notin C\right\}$$
The first condition is satisfied by $b$, $e$, $f$ and $g$, as they do not appear in first position of any pair in $R$. Furthermore, $a$, $c$ and $d$ are not in $C$, and thus satisfy neither of the two conditions, such that we can exclude them from the set. Therefore:
$$T_2=\Delta-\left\{b,e,f,g\right\}=\left\{a,c,d\right\}$$
