I am learning about relations right now and I have a question about some terms.
I am told a relation on $A$ is a subset of $A\times A$. Then I am told a relation $R$ on $A$ is reflexive if for all (and this is where I have a problem) $a\in R$, we have $a\sim a$. Ok sure makes sense, but wouldn't the elements of $R$ be ordered pairs? Then does $ (a,b) \sim (c,d)$ even make sense? I am asked to prove if we have two relations $R$ and $S$ and they are both reflexive then their intersection is reflexive. So do I take an arbitrary element in their intersection and shows its in a relation with itself, but again these are ordered pairs that are not components of other ordered pairs... and I am confused.
No, it's for all $a\in A$ that $a\sim a$. Which means that $(a,a)\in R$
Take, for example, $A=\{1,2,3\}$ then a reflexive relation of $A$ is one that contains all of $(1,1), (2,2), (3,3)$. It may contain other pairs from $A\times A$, but it cannot contain less.
So you have a minimum list of elements that are in any reflexive relation (of $A$), by definition. If all of these elements are in the intersection then that too is a reflexive relation (of $A$).
$R$ is reflexive so: $\forall a\, (a\in A\to (a,a)\in R)$
$S$ is reflexive so: $\forall a\, (a\in A\to (a,a)\in S)$
Can you use this to show that the following must be true? $\forall a\, (a\in A\to (a,a)\in (R\cap S))$.