Q:

Transitive Closure of Symmetric relation

I'm trying to figure out the Transitive Closure of T.  The problem is that I cannot find anything which satisfies (x,y) (y,z) to give (x,z).  Any pointers appreciated. 
$$R,S,T:0..10 \iff 0..10 $$
$$T = R \circ S $$
$$ R = \{(0,8),(1,7),(2,6),(3,5),(5,3),(6,2),(7,1),(8,0)\}$$
$$S = \{(0,8),(1,7),(2,6),(3,5)\} $$
I've calculated T using:
$$R \circ S = \{x \colon R, y \colon S \mid x.2 = y.1 \bullet (x.1,y.2)\}$$
Therefore:
$$T = \{(5,5),(6,6),(7,7),(8,8)\}$$
Reflexive Closure:
$$T \cup \{(0,0),(1,1),(2,2),(3,3),(4,4),(9,9),(10,10)$$
Symmetric Closure:
$$\{(5,5),(6,6),(7,7),(8,8)\}$$
Transitive Closure:
$$???$$

A:

If there are no elements $x,y,z$ with $(x,y) \in T$ and $(y,z) \in T$, then $T$ is already transitive.  This is a case of vacuous truth: The definition of transitivity $$(x,y) \in T, (y,z) \in T \rightarrow (x,z)\in T$$ is automatically satisfied if the left side can never hold.

