Is a relation given by {(a,a), (b,b), (c,c), (a,b), (a,c)} on set {a,b,c} a partial order?

Question

Is was a TRUE/FALSE question in a discrete math exam. I answered False because I believe the relation is neither transitive or not transitive so the definition for a partial order doesn't apply (relation that is reflexive, anti-symmetric, and transitive). I would like a clarification if possible as the exam correction says it is.

Answer 1

It is a partial order. We just need to check the three conditions:

$\underline{\text{Reflexivity}}$: we have $(a,a)$, $(b,b)$ and $(c,c)$, so it is reflexive.

$\underline{\text{Anti-symmetricity}}$: note that we don't have $(x,y)$ and $(y,x)$ for all $x,y \in \{a,b,c\}$, so this condition is trivially satisfied, so it is anti-symmetric.

$\underline{\text{Transitivity}}$: it is easy to check that this condition is satisfied, too. Let's do this carefully: $$(a,a) \ \text{and} \ (a,b) \rightarrow (a,b)$$ $$(a,a) \ \text{and} \ (a,c) \rightarrow (a,c)$$ $$(a,b) \ \text{and} \ (b,b) \rightarrow (b,b)$$ $$(a,c) \ \text{and} \ (c,c) \rightarrow (c,c)$$ This is enough to show that the relation is transitive. Why? All other possible combinations to check don't appear in our relation; for example $(a,b)$ and $(b,c)$ would need to imply $(a,c)$, but $(b,c)$ is not in our relation, so this condition is trivially satisfied.

Answer 2

The relation is clearly reflexive, because all pairs $(a,a), (b,b)$ and $(c,c)$ lie in the relation. It is also antisymmetric: to see this, consider all pairs $(a,b)$ with $a\neq b$ and check that $(b,a)$ is not a member of the relation. In your case, the only pairs of that form are $(a,b)$ and $(a,c)$ and you can see that neither $(b,a)$ nor $(c,a)$ lie in the relation. Hence it is antisymmetric. As for transitivity, you have to consider "composable" pairs of pairs, that is pairs of the form $(a,b)$ and $(b,c)$ where the second component of the first and the first component of the second coincide, and then check that also $(a,c)$ is an element of the relation. In you case, the only such pairs are $(a,a),(a, b)$ and $(a, a),(a, c)$, and in both cases you have that the "composite" pair is inside the relation, namely $(a,b)$ [which is the "composite" of $(a,a)$ and $(a,b)$] and $(a,c)$ [which is the "composite" of $(a,a,)$ and $(a,c)$]