I'm trying to prove the following exercise, but Im a little bit confused:
Let $R$ by a partial order in $A$. Let $B \subseteq A$ then it verifies that $a$ is minimum of $B$ in $R^{-1}$ if and only if $a$ is maximum of $B$ in $R$.
My definitions are:
$1)$ $b \in B$ is minimum of B in $\leqslant$, if $\forall x ∈ B, b \leqslant x$ and
$2)$ $b \in B$ is maximum of B in $\leqslant$, if $\forall x ∈ B, x \leqslant b$.
Proof.
We will denote $R$ as $\leqslant$ and $R^{-1}$ as $\geqslant$.
$(\Longrightarrow).$ Lets supose that $b$ is minimum of $B$ in $\geqslant$, that is, $\forall x\in B$, $x\geqslant b$ this implies, by definition of inverse relation, that $x \leqslant b$ which means that $b$ is maximum of $B$ in $\leqslant$.
$(\Longleftarrow).$ Lets supose that $b$ is maximum of $B$ in $\leqslant$, that is, $\forall x\in B$, $x \leqslant b$ this implies, by definition of inverse relation, that $x \geqslant b$ which means that $b$ is minimum of $B$ in $\geqslant$.
This is my proof and my doubt is in the following fact, given $R^{-1} = \{(y, x): (x, y) \in R\}$ and since Im denotating $R$ as $\leqslant$ and $R^{-1}$ as $\geqslant$ I have in mind that $$y \geqslant x \Longleftrightarrow (y, x) \in R^{-1} \Longleftrightarrow (x, y) \in R \Longleftrightarrow x \leqslant y$$ noting that I am changing the sign of the relation and the order of the variables while in the proof I only change the sign of the relation but not the order of the variables.