Cross Product of Partial Orders im going to have a similar questions on my test tomorrow. I am really stuck on this problem. I don't know how to start. Any sort of help will be appreciated. Thank you
Suppose that (L1;≤_1) and (L2;≤_2) are partially ordered sets. We define a partial order ≤on the set L1 x L2 in the most obvious way- we say (a,b)≤(c,d) if and only if a≤_1 c and b≤_2 d
a)Verify that this is a partial order. Show by example that it may not be a total order.
b)Show tha if (L1;≤_1) and (L2;≤_2) are both lattices, ten so is (L1 x L2;≤).
c)Show that if (L1;≤_1) and (L2;≤_2) are both modular lattices then so is (L1 x L2;≤)
d)Show that if (L1;≤_1) and (L2;≤_2) are both distributive lattices then so is (L1 x L2;≤)
e)Show that if (L1;≤_1) and (L2;≤_2) are both Boolean alegbras, then so is (L1 x L2;≤)

I will talk about part a, then you should give the other parts a try. They will follow in a similar manner (i.e. breaking $\leq$ into its components $\leq_1$ and $\leq_2$).
To show $(L_1 \times L_2, \leq)$ is a partial order, we need to show it is reflexive, anti-symmetric, and transitive.
Reflexivity: Given any $(x,y) \in L_1 \times L_2$ we want to show $(x,y) \leq (x,y)$. Looking at the definition of $\leq$, we are really asking whether both $x \leq_1 x$ and $y \leq_2 y$, which is true since $(L_1, \leq_1)$ and $(L_2, \leq_2)$ are both partial orders.
Anti-symmetry: Suppose $(a,x) \leq (b,y)$ and $(b,y) \leq (a,x)$. This tells us a lot of information:


*

*From the first coordinates, we see $a \leq_1 b$ and $b \leq_1 a$, so $a = b$ (since $\leq_1$ is anti-symmetric).

*From the second coordinates, we see $x \leq_2 y$ and $y \leq_2 x$, so $x = y$ (since $\leq_2$ is anti-symmetric).


Combining those two observations gives $(a,x) = (b,y)$, so $\leq$ is anti-symmetric.
Transitivity: Suppose $(a,x) \leq (b,y)$ and $(b,y) \leq (c,z)$. We want to show $(a,x) \leq (c,z)$. Our hypotheses give lots of information again:


*

*From the first coordinates, we see $a \leq_1 b$ and $b \leq_1 c$. Since $\leq_1$ is transitive, we know $a \leq_1 c$.

*From the first coordinates, we see $x \leq_2 y$ and $y \leq_2 z$. Since $\leq_2$ is transitive, we know $x \leq_2 z$.


Combining those two observations gives $(a,x) \leq (c,z)$, so $\leq$ is transitive.
