# Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both modular lattices then so is $(L_1 \times L_2;≤)$ [duplicate]

Possible Duplicate:
Cross Product of Partial Orders

Suppose that $(L_1;≤_1)$ and $(L_2;≤_2)$ are partially ordered sets. We define a partial order $≤$ on the set $L_1 \times L_2$ in the most obvious way - we say $(a,b)≤(c,d)$ if and only if $a≤_1 c$ and $b≤_2 d$.

a) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both lattices, then so is $(L_1 \times L_2;≤)$.

b) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both modular lattices, then so is $(L_1 \times L_2;≤)$.

c) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both distributive lattices, then so is $(L_1 \times L_2;≤)$.

d) Show that if $(L_1;≤_1)$ and $(L_2;≤_2)$ are both Boolean algebras, then so is $(L_1 \times L_2;≤)$.

I am having trouble solving this problem. Any help would be appreciated.

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Well, show us what you did on (a) and where you got stuck. For example, what is the definition of "lattice". –  GEdgar Oct 11 '12 at 21:14

## marked as duplicate by Henning Makholm, fpqc, tomasz, Thomas, NorbertOct 13 '12 at 17:40

I've been trying to attempt this solution. Here is where I get stuck:

c) $L_1$ and $L_2$ are modular, so

$$L_1: a_1 < a_3 \rightarrow a_1 \vee (a_2 \wedge a_3) = (a_1 \vee a_2) \wedge {a_3}$$

$$L_2: b_1 < b_3 \rightarrow b_1 \vee (b_2 \wedge b_3) = (b_1 \vee b_2) \wedge {b_3}$$

$$L_1\times L_2: (a_1,b_1) < (a_3,b_3) \rightarrow (a_1,b_1) \vee ((a_2,b_2) \wedge (a_3,b_3)) = ((a_1,b_1) \vee (a_2,b_2)) \wedge (a_3,b_3)$$

For $L_1\times L_2$ we have $$(a_1,b_1) < (a_3,b_3) = a_1 < a_3 \text{ and } b_1 < b_3$$

I'm not too sure how to use all this information to prove that $L_1\times L_2$ is a modular lattice. Any guidance would be greatly appreciated.

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