Prove that

length(P x Q) = length(P) + length(Q)

where P, Q are ordered sets of finite length.

Any ideas of where to start...

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- Oct 4th 2011, 11:28 PMGokulength of Lattices
Prove that

**length(P x Q) = length(P) + length(Q)**

where P, Q are ordered sets of finite length.

Any ideas of where to start... - Oct 5th 2011, 11:21 AMemakarovRe: length of Lattices
What is the definition of the length of an ordered set? Are we talking about partial or total orders?

- Oct 5th 2011, 01:37 PMGokuRe: length of Lattices
partially ordered sets, where the length is the the size of the longest chain in the poset.

- Oct 5th 2011, 02:10 PMemakarovRe: length of Lattices
If $\displaystyle a_1,\dots,a_n$ and $\displaystyle b_1,\dots,b_m$ are chains in P and Q, respectively, then $\displaystyle (a_1,b_1),(a_2,b_1),\dots,(a_n,b_1),$$\displaystyle (a_n,b_2),\dots,(a_n,b_m)$ is a chain in P x Q, so length(P x Q) >= length(P) + length(Q). Conversely, if you have a chain $\displaystyle (a_1,b_1)>\dots>(a_n,b_n)$ in P x Q, then for each $\displaystyle i$ you have $\displaystyle a_{i+1}>a_i$ or $\displaystyle b_{i+1}>b_i$ (or both). So, you can construct chains in P and Q whose total length is >= n. This means that length(P) + length(Q) >= length(P x Q).