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2. The main theorem

Theorem. If $ P$ is well partially ordered, then so is $ P ^ *$, the set of finite words (finite sequences) from $ P$, where $ p_1 p_2 \cdots
p_k \leq q_1 q_2 \cdots q_n$ means that $ p_1 \leq q_{i_1}$, $ p_2
\leq q_{i_2}$, ..., $ p_k \le q_{i_k}$ for some $ i_1 < i_2 <
\cdots < i_k$.

Note that Example (c) above is the case where $ P$ is totally unordered.

Some useful notation: Let us write $ p,q,\ldots$ for elements of $ P$ and $ {\bf s}, {\bf t}, \ldots$ for elements of $ P ^ *$. The empty sequence is allowed. Let $ P_{\not\geq q,\geq r} $ mean $ \{ p \in P : p \not\geq q, p \geq r \} $, let $ (P ^ *)_{\geq {\bf t}} $ mean $ \{ {\bf s} \in P ^ * : {\bf s} \geq
{\bf t}\}$, etc. Observe that $ P_{\geq q} $ is an upset of $ P$ and $ P_{\not\geq p}$ is a downset. Also observe that for $ q \in
P$, $ (P _ {\not\geq q}) ^ *$ is the same thing as $ (P ^ *)_{\not\geq q}$ (where $ q$ is regarded as a one-letter word).



Proof of the Theorem.

Suppose $ P ^ *$ is not wpo. Let $ D$ be a downset of $ P$ minimal with respect to the property that $ D ^ *$ is not wpo; by Observation 5, such a $ D$ exists. Without loss of generality, $ D$ is $ P$. Then for all $ q \in
P$, $ (P _ {\not\geq q}) ^ *$, or equivalently $ (P ^ *)_{\not\geq q}$, is wpo. In contrast, $ (P ^ *)_{\not\geq {\bf s}}$ is not wpo for some string $ {\bf s}
\in P ^ *$ in place of $ q$; indeed, let $ {\bf s}$ be any element of an infinite antichain or strictly decreasing sequence in $ P ^ *$.

Re-choose $ {\bf s}$ if necessary to have the least possible length. Write $ {\bf s} = p_1 p_2 \cdots p_k$. Clearly $ {\bf s}$ is not empty, so $ k \geq 1$. Let $ {\bf s}'$ be the shorter string $ p_1 p_2 \cdots p_{k-1}$ (empty if $ k = 1$). Then $ (P ^ *)_{\not\geq {\bf s}'}$ is wpo.

Now examine $ (P ^ *)_{\not\geq {\bf s},\geq {\bf s}'}$. Let $ {\bf t}$ be one of its elements. Since $ {\bf t} \geq {\bf s}'$, we can write $ {\bf t} = {\bf t}_1 q_1 {\bf t}_2 q_2 \cdots {\bf t}_{k-1}
q_{k-1} {\bf t}_k$, where $ q_i \geq p_i$. In fact, by choosing $ {\bf t}_1, {\bf t}_2, \ldots, {\bf t}_{k-1}$ as short as possible in turn (so that $ {\bf t}_1$ stops just before the first element that is $ \geq p_1$, etc.), we have $ {\bf t}_i \not\geq p_i$, for all $ i < k$. Since $ {\bf t} \not\geq {\bf s}$, we have $ {\bf t}_k \not\geq p_k$ as well. In other words, $ (P ^ *)_{\not\geq {\bf s},\geq {\bf s}'}$ is an isotone image of part of $ (P ^ *)_{\not\geq p_1} \times P \times (P ^ *)_{\not\geq
p_2} \times P \times \cdots \times P \times
(P ^ *)_{\not\geq p_k} $, so is wpo. But then $ (P ^ *)_{\not\geq {\bf s}}$, which is the union of the wpo sets $ (P ^ *)_{\not\geq {\bf s},\geq {\bf s}'}$ and $ (P ^ *)_{\not\geq {\bf s}'}$, is also wpo, a contradiction.




next up previous
Next: d_wpo Up: d_wpo Previous: d_wpo
Kirby A. Baker 2003-01-13