Prove $ R \cap (B \times B) $ is partial order on B

Question

Given $R$ is a partial order on $A$ and $B$ is a subset of $A$

Let $x,y,z \in B$

So $x,y,z \in A$

1. So $(x,x) \in R $ Also $(x,x) \in B \times B$. Hence follows the reflexivity.

2. Since $x,y,z \in B$ and so in $A$. So let $xRy$ and $yRz$ so $xRz$ by transititvity of $R$. Also since $x,z \in B$ so $(x,z) \in B \times B$. So $xRz \in R \cap (B \times B)$

3. Now let $xRy$ and $yRx$. This implies $x=y$ because $R$ is partial order on $A$ and $x,y \in A$. Also $(x,y)$ and $(y,x) \in B \times B$. How do I proceed ? Is this even correct?

Thanks

Answer 1

The proof looks complete to me: In 3. you know that if $\left(x,\, y\right),\,\left(y,\, x\right)\,\in\, R\cap\left(B\times B\right)$ they must be in $R$ and thus the antisymmetry directly translates from $A$.

Detailed Proof, as I would write it:

Let $A$ be a set with partial order $R$. Let $B\subseteq A$.

Lemma: Then $R\cap\left(B\times B\right)$ is a partial order on $B$.

Proof:

1.) If $b\in B$ then $b\in A$ and thus $\left(b,b\right)\in R$. Thus $\left(b,b\right)\in R\cap\left(B\times B\right)$. Reflexivity $\checkmark$

2.) If $\left(a,b\right)\in R\cap\left(B\times B\right)$ and $\left(b,c\right)\in R\cap\left(B\times B\right)$ then $\left(a,b\right)\in R$ and $\left(b,c\right)\in R$.

Thus $\left(a,c\right)\in R$ as $R$ is a partial order.

Also from $\left(a,b\right)\in R\cap\left(B\times B\right)$ and $\left(b,c\right)\in R\cap\left(B\times B\right)$ follows that $\left\{ a,\, b,\, c\right\} \subset B$.

Thus $\left(a,c\right)\in R\cap\left(B\times B\right)$. Transitivity $\checkmark$

3.) If $\left(a,b\right)\in R\cap\left(B\times B\right)$ and $\left(b,a\right)\in R\cap\left(B\times B\right)$ then both $\left(a,b\right)\in R$ and $\left(b,a\right)\in R$.

Thus $b=a$ as $R$ is a partial order. Antisymmetry $\checkmark$

qed