$R_1$ and $R_2$ are partial orders. What about $R_1 \cap R_2$?

Question

Let $R_1$ and $R_2$ be two partial order relations defined on a set S. Show that $R_1 \cap R_2$ is also a partial order on S.

I am struggling to represent $R_1$ and $R_2$ in a way I can operate with $\cap$.

Answer 1

$R_1$ and $R_2$ are by definition subsets of $S \times S$ which are reflexive, antisymmetric, and transitive. Now we need to check that $R_1 \cap R_2$ is also reflexive, antisymmetric, and transitive.

• Reflexive: Since $(a,a)$ must be in both $R_1$ and $R_2$ for any $a \in S$, $(a,a)$ will also be in $R_1 \cap R_2$, so it is reflexive.
• Antisymmetric: Now, this is a conditional property. If $(a,b) \in R_1 \cap R_2$ and $(b,a) \in R_1 \cap R_2$, it must be the case that $a = b$. Since we know that this property is satisfied for both $R_1$ and $R_2$, it must also hold for $R_1 \cap R_2$.
• Transitivity: Likewise, since we know that the transitive property holds for both $R_1$ and $R_2$, there can be no two elements $(a,b) \in R_1 \cap R_2$ and $(b,c) \in R_1 \cap R_2$ without it also being the case that $(a,c) \in R_1 \cap R_2$.

Answer 2

• reflexivity

$\forall x \in S$, we have $(x, x) \in R_1$ and $(x, x) \in R_2$, thus $$ (x, x) \in R_1 \cap R_2 $$

• antisymmetry

if $(x,y) \in R_1 \cap R_2$ and $(y, x) \in R_1 \cap R_2$, then $$ (x,y) \in R_1 \text{ and } (y, x) \in R_1 $$ thus $x = y$.

• transitivity

if $(x, y) \in R_1 \cap R_2$ and $(y, z) \in R_1 \cap R_2$, we have $$ (x,y)\in R_1, (y,z)\in R_1 \implies (x, z) \in R_1 \\ (x,y)\in R_2, (y, z)\in R_2 \implies (x, z) \in R_2 $$ thus $(x,z) \in R_1 \cap R_2$.