How to prove that for any partial order $R$ on $A$ exists $S$ partial order on $A$ so that $R$ is a subset of $S$. $S \ne R.$
I have tried to create a new relation S which is A Union of R with a pair $(a,b)$ that $(a,b),(b,a)$ not belong to R. $(a,b)\in $ A. but its become a little bit to complicated to make $S$ be a partial order.
thaks
On
I’m not sure what your definition of a partial order is, but this is false for any total order. Take $(\mathbb{R}, \leq)$ as an (counter-)example. Suppose there exists a partial order $R$ that strictly contains $\leq$. Then there is a pair of real numbers $(a,b)\in R$ that is not in $\leq$. This is only possible if $a>b$ and $(b,a)\in \phantom{}\leq$. Since $R$ contains $\leq$, it follows that $(b,a)\in R$ too. Recall that $(a,b)\in R$, therefore by anti-symmetry we have that $a=b$. But this contradicts our assumption that $(a,b)\notin \phantom{}\leq$, so no such pair $(a,b)$ exists. (Excuse the clumsy notation, but I feel this makes the proof easier to follow)
If $A=\{0\}$ and $R=\{(0,0)\}$ you obviously can't construct a strictly bigger partial order on $A$.