# Partially ordered set and injection functions

Suppose $<A,\prec>$ is partially ordered set and $f(x)=${$a\in A|a\prec x$} is a function. Prove that $f$ is not injection function: there are $y \neq z \in A$ such as $f(y)=f(z)$.

While I can pretty easily prove this statement for a specific partially ordered set, I somehow cannot find a formal prove that will be valid for any partially ordered set.

Thank you.

• That doesn't look true to me. – Lord Shark the Unknown Sep 30 '18 at 6:20
• any well ordered set is partially ordered, yet such a mapping is injective. If I get this right the mapping is $A\to\mathcal P(A), x\mapsto \{a< x\}$ – Alvin Lepik Sep 30 '18 at 6:52

If your function collects elements strictly greater than the given element (so $$f(a)$$ doe not contain $$a$$) your function can not be injective over some set; for example, consider the set $$A=\{1,2,3,6\}$$ and the divisibility relation $$\mid$$. You can see that $$f(2)=f(3)=\{6\}$$.
If $$f(a)$$ contains $$a$$, then $$f$$ is injective; if $$f(a)=f(b)$$ then $$a\in f(b)$$ and $$b\in f(a)$$, so $$a\le b$$ and $$b\le a$$.