I need to prove that the intersection of 2 POsets R and S is a POset.
So we basically want to prove that if $R$ and $S$ are POsets then $R \cap S$ is reflexive, transitive and anti-symmetric.
The problem is in the reflexive, I proved it like this:
We need to prove that $\forall a \in A ~~~ <a,a> \in R \cap S$
Because $R$ and $S$ are POsets, if $R \cap S = \emptyset$ then the intersection is reflexive (empty-wise)
Else, because $R$ and $S$ are POsets, if $<a,b> \in R \cap S$ then $<a,a> \in R$ and $<b,b> \in S$ because they are reflexive.
And thus $<a,a> , <b,b> \in R \cap S$ and thus it is reflexive.
In the comments the professor said: Reflexive is not correct, you wrote that one thing is in R and other thing is in S so it is in the intersection (?) (-4 pts)
I don't know why I am wrong here, I would appreciate if you could point me to the mistake
Note: this does not have to be 'well written', we are first year students..I just need to find the logical mistake, not the semantic.
Thank you!
Hint: from the comments, there is no constraint on the underlying sets of the posets $R$ and $S$. As you have noted, you can prove reflexivity and antisymmetry of $R \cap S$ without constraining the underlying set (or field) of this relation. For reflexivity, what you have to prove is that for every $x$ in the field of the relation $R \cap S$, $\langle x, x \rangle \in R \cap S$. Your argument is right if you define $A$ to be the field of $R \cap S$ (although you might want to improve the phrasing - you have correctly identified the conditions for $a$ or $b$ to be in the field of $R \cap S$). I don't think your professor's comments are very helpful: the right comment is "what's $A$?". Maybe he or she thought they'd asked a question about posets $R$ and $S$ over the same field, but, apparently, they did not.