TITLE: $(\forall x \in A)(\forall y \in A)(x Ry \iff P_x ⊆ P_y)$ for a partial order $R$
QUESTION [1 upvotes]: This is Velleman's exercise 4.4.10:
Suppose $R$ is a partial order on $A$. For each $x ∈ A$, let $P_x = \{a ∈ A | aRx\}$. Prove that $∀x ∈ A∀y ∈ A(x Ry ↔ P_x ⊆ P_y)$.
Here's my proof of it:
Proof. Suppose $x$ and $y$ to be arbitrary elements of $A$.
($\rightarrow$) Suppose $xRy$. Suppose $aRx$. Since $R$ is transitive then from $aRx$ and $xRy$ we get $aRy$. Therefore if $aRx$ then $aRy$ and therefore if $xRy$ then $(aRx \rightarrow aRy).$
($\leftarrow$) Suppose $(aRx \rightarrow aRy)$. By the definition of partial order we have $xRy$ and therefore if $(aRx \rightarrow aRy)$ then $xRy$.
From ($\rightarrow$) and ($\leftarrow$) we'll have $(x Ry ↔ P_x ⊆ P_y)$ and since $x$ and $y$ were arbitrary then $∀x ∈ A∀y ∈ A(x Ry ↔ P_x ⊆ P_y)$.
Is my proof correct? Particualrly the backward direction in which I didn't use "$(aRx \rightarrow aRy)$" directly. 
Thanks a lot.

REPLY [3 votes]: The $\leftarrow$ does indeed need a bit of work.
Use that $R$ is reflexive to get that $xRx$, and hence $x \in P_x$. Since $P_x \subseteq P_y$ that means that $x \in P_y$, and hence $xRy$

REPLY [2 votes]: It seems correct, but you did skip a step in the backward direction with the words "by the definition of partial order". Here's a rewording:
Suppose $P_x\subseteq P_y$. Since $R$ is reflexive, we have $x\in P_x$. Thus the assumption gives $x\in P_y$, so that $xRy$.
This shows that you do infact use $(aRx\to aRy)$.
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