Given two arbitrary Orderings $R$ and $S$ on an arbitrary set $X$, decide which of the following will necessarily be an ordering. a) $R\cap S$ b) $R \cup S$ c) $R\circ S$ (concatenation of R and S).
For a) I've argued that $R \cap S$ is only applicable to subset of $X$, hence it is also an ordering.
b) I've said that we are just taking a union of two ordering without changing any properties of the ordering, so it must also be an ordering.
c) I am not so sure about c. Would appreciate any help.
I suppose with orderings you mean partial orderings, that is, binary relations which are reflexive, anti-symmetric and transitive.
So let $R$ and $S$ be two such relations on a set $X$.
(a) It's clear that that $R \cap S$ is reflexive; if $(a,b), (b,a) \in R \cap S$, then they belongs to both pairs, whence $b=a$ using the anti-symmetry in one of the relations (anyone); for transitivity $(a,b), (b,c) \in R \cap S$, then these pairs belong to each of the relations, which are transitive, whence $(a,c)$ belongs to each of them too and therefore to $R \cap S$.
(b) $R \cup S$ doesn't have to be a partial ordering.
For example, say $X = \{0,1\}$, $R = \{(0,0),(0,1),(1,1)\}$ and $S = \{(0,0),(1,0),(1,1)\}$.
Then $(0,1), (1,0) \in R \cup S$; by anti-symmetry it should be $0=1$, which is not the case.
(c) $R \circ S$ doesn't have the be partial ordering either.
Take the same example as in (b). From $(0,1) \in R$ and $(1,1) \in S$, it follows that $(0,1) \in R \circ S$; from $(1,1) \in R$ and $(1,0) \in S$, it follows that $(1,0) \in R \circ S$; again from this it should follow that $0=1$, if this was a partial ordering. Hence it is not.