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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;≤)

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Where to start when you're asked to show that this-or-that is a something, is to find the definition of something and write out what exactly it means for a this-or-that. In case (a) here, find the definition of "partial order", and write out what that means in the particular case of $L_1\times L_2$. Similarly for all of the other questions with something being "lattice", "modular lattice", "distributive lattice" and "Boolean algebra". –  Henning Makholm Oct 11 '12 at 20:23
    
Thanks for that, but I'm having a lot of difficulty with the next few answers. Could you possible give direction for c or d? –  user44438 Oct 12 '12 at 1:43

1 Answer 1

up vote 3 down vote accepted

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.

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