# Partial orders

• May 5th 2010, 02:18 PM
TheFangel
Partial orders
Hi,

I have to prove that $R \cap (BxB)$ is a partial order on B, supposed that R is a partial order on A and $B \subseteq A$.
• May 6th 2010, 12:34 AM
Swlabr
Quote:

Originally Posted by TheFangel
Hi,

I have to prove that $R \cap (BxB)$ is a partial order on B, supposed that R is a partial order on A and $B \subseteq A$.

You have three things to prove, and writing $(a, b) \in R$ for $a \leq b$ or $a R b$ or whatever,

$(a, a) \in R$

$(a, b) \in R \text{ and }(b, a) \in R \Rightarrow a=b$

$(a, b) \in R \text{ and } (b, c) \in R \Rightarrow (a, c) \in R$.

You know this holds for all $a, b, c \in A$ and you need to show that it holds for all $a, b, c \in B \subseteq A$. However, as $a, b, c \in B \Rightarrow a, b, c \in A$ then...