Union and intersection of two different relations

Question

Let's say I have a relation $\mathcal{R}$ that is equivalence (reflexive, transitive and symmetrical) and relation $\mathcal{S}$ that is partial order (reflexive, transitive and antisymmetrical).

How can I find out what does their union $(\mathcal{R}∪S)$ and their intersection $(\mathcal{R}\cap \mathcal{S})$ will be like? I.e. whether it also must be symmetrical, reflexive (etc.) or not?

Answer 1

Reflexivity holds for both $\mathcal{R} \cup \mathcal{S}$ and $\mathcal{R} \cap \mathcal{S}$, obviously.

For symmetry, it is inconclusive I think. For example take $A = \{1,2,3\}$ and $\mathcal{R} = (A, \{(1,1),(2,2),(3,3),(1,2),(2,1)\})$ and $\mathcal{S} = (A, \{(1,1),(2,2),(3,3),(1,2)\})$. Now $\mathcal{R} \cup \mathcal{S} = \mathcal{R}$ (equivalance relation) and $\mathcal{R} \cap \mathcal{S} = \mathcal{S}$ (partial ordering). But if we take $\mathcal{R} = (A, \{(1,1),(2,2),(3,3),(1,2),(2,1)\})$ and $\mathcal{S} = (A, \{(1,1),(2,2),(3,3),(1,3)\})$, we have $\mathcal{R} \cup \mathcal{S} = (A, \{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3)\})$ (neither equivalance relation nor partial ordering) and $\mathcal{R} \cap \mathcal{S} = (A, \{(1,1),(2,2),(3,3)\})$ (equivalance relation).

For transitivity, it is inconclusive for $\mathcal{R} \cup \mathcal{S}$ by the example given in symmetry. It holds for $\mathcal{R} \cap \mathcal{S}$ because if you define a relation $\mathcal{T} = \mathcal{R} \cap \mathcal{S}$, this is equivalent to the following statement:

$x \mathcal{T}y$ if and only if $x \mathcal{R} y$ and $x \mathcal{S} y$. Now, we know that transitivity holds for both $\mathcal{R}$ and $\mathcal{S}$. So for all $x,y,z$ we have $$(x\mathcal{R}y \land y\mathcal{R}z \land x\mathcal{R}z) \land (x\mathcal{S}y \land y\mathcal{S}z \land x\mathcal{S}z) = (x\mathcal{R}y \land x\mathcal{S}y) \land (y\mathcal{R}z \land y\mathcal{S}z) \land (x\mathcal{R}z \land x\mathcal{S}z)$$ (In this notation, $x\mathcal{R}y$ means $x$ is related to $y$) which is equivalent to $$x\mathcal{T}y \land y\mathcal{T}z \land x\mathcal{T}z$$ So $\mathcal{T}$ is transitive.

Answer 2

Relations with a property of the form
. . for all $a,b,c,... P(a,b,c,...)$ for elements $a,b,c,...$ are closed under intersection.

Relations with a property of the form
. . for all $a$, $P(a) $ for elements $a$, are closed under union.

Usually for unions, these generalizations are inadequent.