# Question about Partially Ordered Sets and functions

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 ?

$$h(x)=h(y)$$ $$\Longrightarrow h(x)\preceq^{\prime}h(y)\:\land\:h(y)\preceq^{\prime}h(x)$$ $$\Longrightarrow x\preceq y\:\land\:y\preceq x$$ $$\Longrightarrow x=y$$
• No need to work with a contradiction: $h(x) = h(y)$ implies $x = y$ as you showed, so $h$ is one-to-one. – Mark Kamsma Apr 16 '19 at 21:29