discrete math binary relations - why is set transitive

Question

The question is:

Let $R$, $S$ be partial orders defined on $A$. Is $R \cap S$ a partial order too? Justify.

The school answer given states that $R$ and $S$ are transitive since if $(x,y), (y,z) \in R \cap S$, then $(x,y), (y,z) \in R$ and $(x,y), (y,z) \in S$. Hence, $(x,z) \in R ∩ S$.

In my answer, I gave an example on transitivity seen in $R$ and $S$ respectively, but transitivity not observed in $R ∩ S$. Why is it wrong?

Answer 1

The only pairs $(a,b)$ and $(b,c)$ that would be useful to refute transitivity of $R\cap S$ would be those where $(a,b),(b,c)\in R\cap S$, but $(a,c)\not\in R\cap S$.

In your example, $(2,4)\in R$ but $(2,4)\not\in S$ so $(2,4)\not\in R\cap S$. Similarly, $(4,5)\in S$ but $(4,5)\not\in R$, therefore $(4,5)\not\in R\cap S$. This means those two pairs are not useful in determining (non)-transitivity. And, actually $R\cap S$ is transitive because $R\cap S=\{(3,3),(4,4)\}$.

I am also completely at loss about why you claim $R$ and $S$ are transitive in the first place. $(2,4), (4,3)\in R$ but $(2,3)\not\in R$, and, similarly, $(3,4),(4,5)\in S$ but $(3,5)\not\in S$.