Generally, the mistake that we do with statement S1 is that, we assume that since $[N,|]$ is a poset, so is the $[R,|]$. But what we fundamentally miss is, that the divisibility relation in $N$ satisfies the antisymmetric property. Whereas, $R$ doesn’t provide antisymmetric property with the divisibility relation. This is also the case with $Z,Q,Q’$ with the divisibility relation.
Ex: (2,-2) and (-2,2) both belong to the relation $[R,|]$. Hence, not a Poset.
Question: 1. If $A\subset B$ and $(A,\preceq)$ is a poset then $(B,\preceq)$ is also a Poset?
- If $A\subset B$ and $(B,\preceq)$ is a poset then $(A,\preceq)$ is also a Poset?
S2: If a relation $R$(represents a relation and not the set of real numbers) on a set $A$ is irreflexive and transitive then $R$ is antisymmetric.
To check if the S2 is true we just need to check if the relation $R$ is antisymmetric. Explicitly, we need to check if both $(a,b)\in R$ and $(b,a)\in R$, with $a\neq b$. If they do then $R$ is not anti-symmetric, else $R$ is antisymmetric.
Let, $a\neq b$ . Assume that, $(a,b)\in R$ and $(b,a)\in R$. Then, by $transitivity$ of relation $R$, we have that $(a,a)\in R$. But, this is a contradiction, as $R$ is $irreflexive$. Thus, our assumption was wrong.
Hence, $R$ is $antisymmetric$. So, S2 is True.