finite sequences,linear order, lexicographic order, well order, cardinal

Question

Let $\lambda$ be an infinite cardinal.Consider the lexicographic order on $\lambda^{< \omega}$. Why this order is not a well-order: how may an infinite long descending chain look like? Why every descending chain is of order type at most $\omega$? Is it true that in lexicographic order a shorter sequence precedes a longer one, and then the order of these two is decided componentwise ?

Answer 1

The lexicographic order $\preceq$ on sequences of various lengths is indeed defined in such a way that if $\sigma$ is an initial segment of $\tau$, then $\sigma\preceq\tau$, and in all other cases the order is decided by the first position at which $\sigma$ and $\tau$ differ. That is not why there is an infinite descending chain in $\lambda^{<\omega}$, however: none of the sequences in the following chain is an initial segment of any of the others.

$$1\succ01\succ001\succ0001\succ00001\succ\ldots$$

With this as a model you should be able to form many other infinite descending chains.

It’s not true that every descending chain is of order type at most $\omega$. Consider the descending chain

$$2\succ12\succ112\succ1112\succ11112\succ\ldots$$

of order type $\omega$; can you find a way to extend it to a descending chain of type $\omega+1$? How about $\omega+\omega$?