Hi,

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

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- May 5th 2010, 02:18 PMTheFangelPartial orders
Hi,

I have to prove that $\displaystyle R \cap (BxB)$ is a partial order on B, supposed that R is a partial order on A and $\displaystyle B \subseteq A$. - May 6th 2010, 12:34 AMSwlabr
You have three things to prove, and writing $\displaystyle (a, b) \in R$ for $\displaystyle a \leq b$ or $\displaystyle a R b$ or whatever,

$\displaystyle (a, a) \in R$

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

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

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