# Can this relation be transitive but not symmetric and reflexive?

Let $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$. Give an example of a relation $T$ on $A$ with at least three elements that is not reflexive, not symmetric, but transitive. Explain clearly why your solution has/doesn’t have the given properties.

I am stuck on this question as to be transitive i thought you needed to have $(x,y), (y,x)$ and $(x,x)$.

So i have worked out the relation as $T = \{(1,1), (1,2), (2,1)\}$.

So for example $(1,2), (2,1)$ and $(1,1)$ makes it transitive.

and, from what I understand, it is reflexive as $(x,x)$ are elements of $T$ if all $x$ is a elements of $A$.

It is symmetric as $x$ and $y$ are elements of $A$ and $(x,y)$ is a element of T and $(y,x)$ is a element of $T$ which means you cannot find such a set?

• Hint: revisit the definition of what it means for a relation to be transitive (which involves three elements of the underlying set, not necessarily distinct) – Mark Bennet May 5 '15 at 15:16
• yes sorry i see my mistake straight away as transitive is (x,y) (y,z) and (x,z)??? is this correct? @MarkBennet – John May 5 '15 at 15:21
• @thomas Yes, that is correct. – Lord_Farin May 5 '15 at 15:31
• How about $<$ on $\{1,2,3\}$? – Hagen von Eitzen May 5 '15 at 15:39

Hint: Since we need to break symmetry, there has to be some $(a,b) \in T$ with $a \ne b$. Let us pick $(1,2) \in T$.
Now we need to add some extra things to $T$ to satisfy the requirement that it have three or more elements. (Recall that $T$ is transitive iff $(x,y), (y,z) \in T \implies (x,z) \in T$.)
Suppose that we add $(2,3)$ to $T$. What is missing from $T$ to make it transitive? Is the result reflexive? Symmetric? Transitive?