Defining left shift on orderings

Question

The question I propose is this: For an indexing set $I = \mathbb{N}$, or $I = \mathbb{Z}$, and some alphabet $A$, we can define a left shift $\sigma : A^{I} \to A^{I}$ by $\sigma(a_{k})_{k \in I} = (a_{k + 1})_{k \in I}$, because there exists a unique successor $\min \{ i > i_{0} \}$ for all $i \in I$. But one could not do the same if we made, say, $I = \mathbb{Q}$. Is there a way to characterize this in the language of orderings, i.e. a way to characterize the existence of a unique successor of a directed set such that one could sensibly define a "left shift" on a net $s: I \to A$? Does this place restrictions on, say, the cardinality of $I$?

Answer 1

There's really no issue here, given any partial order $P$, you can take the lexicographic product of $P$ with $\Bbb N$ or $\Bbb Z$, and obtain a partial order where each element has a unique immediate successor.

So there is no limit on the cardinality of such set. If, however, you want every two points to be within a finite distance of successive steps, you have to have it countable, and a subset of $\Bbb Z$.

Answer 2

Let $\langle I,\preceq\rangle$ be a linear order; all that’s required is that each $i\in I$ have an immediate successor $i^+$ in the order $\preceq$: $i\prec i^+$, and there is no $j\in I$ such that $i\prec j\prec i^+$.

Define a relation $\sim$ on $I$ as follows: for $i,j\in I$, $i\sim j$ iff either $i\preceq j$ and $[i,j]$ is finite, or $j\preceq i$ and $[j,i]$ is finite, where the intervals are of course taken with respect to the order $\preceq$. It’s easy to check that $\sim$ is an equivalence relation. For $i\in I$ let $[i]$ be the $\sim$-equivalence class of $i$; it’s also pretty straightforward to check that either $[i]$ has no least element and is order-isomorphic to $\Bbb Z$, or $[i]$ has a least element and is order-isomorphic to $\Bbb N$.

Let $\mathscr{I}=\{[i]:i\in I\}$; $\preceq$ induces a natural linear order $\sqsubseteq$ on $\mathscr{I}$ by $[i]\sqsubseteq[j]$ iff $i\sim j$ or $i\prec j$, and it turns out that $\langle\mathscr{I},\sqsubseteq\rangle$ can be any linear order whatsoever.

To see this, let $\langle L,\le\rangle$ be a linear order, and let $L_0$ be any subset of $L$. Let

$$I=\{\langle x,n\rangle\in L\times\Bbb Z:n\ge 0\text{ if }x\in L_0\}\;,$$

and let $\preceq$ be the lexicographic order on $I$: $\langle x,m\rangle\preceq\langle y,n\rangle$ iff $x<y$, or $x=y$ and $m\le n$. Then

$$[\langle x,n\rangle]=\begin{cases} \{x\}\times\Bbb N,&\text{if }x\in L_0\\ \{x\}\times\Bbb Z,&\text{if }x\in L\setminus L_0\;, \end{cases}$$

and $\langle x,n\rangle^+=\langle x,n+1\rangle$ for each $\langle x,n\rangle\in I$.

In particular, $I$ can be something quite different from a well-order, e.g. $\Bbb Q\times\Bbb Z$ ordered lexicographically.

Up to here I’ve assumed that you want a linearly ordered index set, but it should be clear that $L$ can just as well be a partial order.