If we define 2 reflexive relations(say R and S) a set A, then are not they same?
Reflexive relation means for all $\forall x \in A, (x,x) \in R$. Same is the case for S. Then isn't R = S???
if so we consider the theorem, if R and S are reflexive relations then $R \cup S, R \cap S$ are both reflexive. When we define R and S on same set A, are not they R=S? Then theorem becomes trivial right.
On
No reflexive relations on a set do not have to be the same.
For example consider $$A=\{1,2,3\}$$
$$R= \{(1,1), (2,2),(3,3), (1,3)\}$$ and $$ S= \{(1,1), (2,2),(3,3), (2,3)\}$$ Where both relations are reflexive but they are not the same.
As you notice,the intersecton and the union are reflexive in this example but you have to prove it in general.
On
No, they are not necessarily the same. $\forall x\in A: (x,x) \in R$ does not mean that these are the only pairs in $R$, there can be many more, which in turn may or may not be in $S$.
For example the relations "have the same parity" and "have the same remainder when divided by $3$" on $\mathbb{Z}$ are both reflexive. $(1,3)$ is included in the first but not in the second, while $(0,3)$ is included in the second but not in the first (Their intersection is the relation "have the same remainder when divided by 6").
On
$R$ being a reflexive relation on a set $A$ implies that identity relation $I = \{(x,y):x=y\}$ where $x,y\in A$, is a subset of $R$, $I\subseteq R$. So if two sets $S$ and $T$ are reflexive relations on a set $A$, then we can only conclude that $I\subseteq S$ and $I\subseteq T$. Obviously, $I\subseteq S\cup T$ and $I\subseteq S\cap T$, therefore both intersection and union of the sets $S$ and $T$ are also reflexive.
No. For example, on $\mathbb R$, $=$ and $\le$ are both reflexive relations, but not the same.