Tags: #theorem

Statement

The product of chains has a symmetric chain decomposition.

Proof

Lemma 1: The product of two chains have a SCD
We may construct it as follows

Note that these are symmetric as the first one is and all subsequent $L$'s start at a rank one below on top and one above on the bottom.

Lemma 2: If two finite ranked posets $P,Q$ have an SCD, then $P\times D$ has an SCD too.
Let $P=C_1\bigsqcup C_2\bigsqcup \cdots \bigsqcup C_k$ and $Q=C_1^{\prime }\bigsqcup C_2^{\prime }\bigsqcup \cdots \bigsqcup C_l^{\prime }$ be their respective SCDs. Then, by lemma 1, $C_i\times C_j^{\prime }$ have an SCD for all $i,j$. We claim that the collection of all of these form an SCD for $P\times Q$.

Example:

We gloss over why these chains are still symmetric (product of symmetric is symmetric, roughly).

Then, by induction, any arbitrary product of chains have an SCD.