Finding Reflexive and Transitive Relation

Question

Let $A =\{ 1, 2, 3 \}$

How can I find a relation from $A$ to $A$ that is both reflexive and transitive but not symmetric?

Would the solution be $\{ (1, 1), (2, 2), (3, 3) \}$?

Answer 1

$A = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$

The witness of the non-symmetry is $(1,2) \in A$ but $(2,1) \notin A$.

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Mace4 constructed eight models:

1. $A = \{(1,1),(2,2),(3,3),(1,2)\}$
2. $A = \{(1,1),(2,2),(3,3),(1,2),(3,2)\}$
3. $A = \{(1,1),(2,2),(3,3),(1,2),(3,1),(3,2)\}$
4. $A = \{(1,1),(2,2),(3,3),(1,2),(1,3)\}$
5. $A = \{(1,1),(2,2),(3,3),(1,2),(1,3),(3,2)\}$
6. $A = \{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)\}$
7. $A = \{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,2)\}$
8. $A = \{(1,1),(2,2),(3,3),(1,2),(1,3),(3,1),(3,2)\}$

Answer 2

This is the same answer as the one provided by Kenny Lau, but with a different notation. Consider the relation $\leqslant$ (just the usual “smaller than or equal to”) on $A$. It should be clear that it is reflexive, not symmetric, and transitive.