Could someone verify my attempt at the following problem?
Let $(A,\preceq)$ and $(B, \preceq ')$ be Partially ordered sets and suppose that $h:A \rightarrow B$ satisfies $x \preceq y \iff h(x) \preceq 'h(y)$ for all $x,y \in A$.Prove that h is one-to-one.
Attempt:
Let $(A,\preceq)$ and $(B, \preceq ')$ be POSETS and suppose that $h:A \rightarrow B$ satisfies $x \preceq y \iff h(x)\preceq 'h(y)$ for all $x,y \in A$.
Suppose for a contradiction that $h$ is not one-to-one,then there exists some $x,y \in A$ such that (i) $h(x)=h(y)$ and $x\neq y$.
By anti-symmetry,if $x \preceq y$ and $y \preceq x$ then $x=y$. The logically equivalent contraposition suggests that if $x \neq y$ then $ x \npreceq y$ or $y \npreceq x$.
Since $x \neq y$ by (i) then the following divison by cases are possible.
[case : $x \npreceq y$ and $y \preceq x$]
by $x \preceq y \iff h(x)\preceq 'h(y)$ the following holds :
- $x \npreceq y$ implies $h(x)\npreceq 'h(y)$ and
- $y \preceq x$ implies $h(y)\preceq 'h(x)$
therefore with $h(x)\npreceq 'h(y)$ and $h(y)\preceq 'h(x)$ then $h(x) \neq h(y)$ which contradicts (i) according to which $h(x)=h(y)$
[case : $x \preceq y$ and $y \npreceq x$]
..similar as the previous case
[case : $x \npreceq y$ and $y \npreceq x$]
by $x \preceq y \iff h(x)\preceq 'h(y)$ the following holds :
- $x \npreceq y$ implies $h(x)\npreceq 'h(y)$ and
- $y \npreceq x$ implies $h(y)\npreceq 'h(x)$
therefore with $h(x)\npreceq 'h(y)$ and $h(y)\npreceq 'h(x)$ then $h(x) \neq h(y)$ which contradicts (i) according to which $h(x)=h(y)$.
Therefore h is one-to-one.
it looks right but feels iffy to me because for each of the cases i conclude that $h(x) \neq h(y)$ on the bases that
if $h(x) \preceq 'h(y)$ and $h(y)\preceq 'h(x)$ then $h(x)=h(y)$.
where for example in the last case, $h(x)\npreceq 'h(y)$ and $h(y)\npreceq 'h(x)$ was implied to yield $h(x) \neq h(y)$ ...is that logically right ?
thank you for your time.
You don't really need proof by contradiction
$$h(x)=h(y)$$ $$\Longrightarrow h(x)\preceq^{\prime}h(y)\:\land\:h(y)\preceq^{\prime}h(x)$$ $$\Longrightarrow x\preceq y\:\land\:y\preceq x$$ $$\Longrightarrow x=y$$