Let (A, $\preceq$) be a poset.Define a map $f:A \to P(A)$ by $f(x) = \{ y \in A | y \preceq x \}$ for all $x \in A$.

Question

Let (A, $\preceq$) be a poset.Define a map $f:A \to P(A)$ by $f(x) = > \{ y \in A | y \preceq x \}$ for all $x \in A$. Let $x,y \in A$.Show that $x\preceq y$ iff $f(x) \subseteq f(y)$

Proof:

Firstly, let assume $x\preceq y $.Since (A, $\preceq$) is a poset, we have $y \preceq x$, so by the definition of f, we have $y\in f(x)$.Since $\preceq$, is reflexive, $y\preceq y$, so $y\in f(y)$, which implies $f(x) \subseteq f(y)$.

Secondly, let $f(x) \subseteq f(y)$, and let $y\in f(x)$.From the hypothesis of the argument, we have $y\preceq x$, so by symmetriccity, we have $x\preceq y$.

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So, Is there any problem in the proof, any flow or any point that I should write more clear or use a different style to express ?

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Edit: Ok, I found one, in the first part of the proof, since $(A,\preceq)$ is a poset, it is antisymmetric, not symmetric, so it doesn'T imply $y \preceq x$, which is used also in the second part of the proof, so I hoping you can give me a proper proof.

Answer 1

Your proof has some flaws, but the overall idea is correct.

Let's start assuming $x \preceq y$. You have to prove that $f(x) \subseteq f(y)$. By definition that means that $$ \forall z : z\in f(x) \implies z \in f(y) $$ You cannot restrict yourself to $x$ and $y$. So let $z \in f(x)$, that is, by definition of $f$, $z \preceq x$. As $\preceq$ is transitive, $z \preceq y$, hence $z \in f(y)$. This proves $f(x) \subseteq f(y)$.

For the other direction, let $f(x) \subseteq f(y)$ hold. Then, as by reflexivity, $x \in f(x)$, we have $x \in f(y)$. That implies $x \preceq y$ by definition of $f$.

Answer 2

Suppose that $x \preceq y$. Let $z \in f(x)$, so $z \preceq x$. By transitivity of $\preceq$ we know that $z \preceq y$, so $z \in f(y)$. As $z$ was arbitrary we know that $f(x) \subseteq f(y)$.

Suppose that $f(x) \subseteq f(y)$. Because $x \preceq x$, we know that $x \in f(x)$, so by the inclusion, $x \in f(y)$ which means that $x \preceq y$ by definition.

So the two basic axioms (idempotence and transitivity) of a pre-order give this equivalence $x \preceq y \leftrightarrow f(x) \subseteq f(y)$ straightforwardly. Antisymmetry is not needed.

Answer 3

The $\Rightarrow$ direction:

If $x\leq y$, then for any $z\in f(x)$ by the transitive we have that $z\leq y $ and so $z\in f(y)$. Hence $f(x)\subseteq f(y)$.

The $\Leftarrow$ direction:

If $f(x)\subseteq f(y)$ then as $x\in f(x)$, this implies $x\leq y$ by definition of $f(y)$.