# an infinite queue preserving equality.

Is there any well-ordered set $(A,\leq)$ such that:

1. $(A,\leq^{-1})$ is well-ordered.
2. $A$ is infinite.
3. there's exactly one function $\theta:A\rightarrow \{0,1\}$ such that

1) for each $a < M$, $$\theta(a)=\theta(a^+)$$ 2) for each $b > m$, $$\theta(b)=\theta(b^-)$$

where $$m=\min (A)$$ $$M=\max (A)$$ $$a^+=\min\{x\in A\mid a<x\}$$ $$b^-=\max\{x\in A\mid x<b\}$$

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By $\leq^{-1}$ are you denoting the reverse ordering: $x \leq^{-1} y$ iff $y \leq x$? – arjafi Jan 29 '13 at 12:21
yes.56789012345 – user59671 Jan 29 '13 at 12:29

If $( A , \leq )$ is an infinite well-ordered set, then $( \mathbb{N} , \leq )$ embeds into $( A , \leq )$, that is, there is a one-to-one function $f : \mathbb{N} \to A$ such that $m \leq n \; \Rightarrow \; f(m) \leq f(n)$. Letting $X$ denote the image of this mapping it follows that $X$ has no $\leq^{-1}$-least element (since it has no $\leq$-greatest element), and thus $( A , \leq^{-1} )$ is not well-ordered.
so my simulation is wrong. however $1,2,3,...,-3,-2,-1$ may be the best simulation. – user59671 Jan 29 '13 at 12:38
I needed well ordering for $a^+$ and $b^-$. – user59671 Jan 29 '13 at 12:43
I honestly no idea what you mean by your simulation. Perhaps it would be best to figure out exactly what you want to be modelling. At the very least it seems that you want an infinite ordered set with a first (least) element, and such that every element (except possibly a greatest element) has an immediate successor. It also appears that you want everything except the first element to have an immediate predecessor. Why is $\mathbb N$ not appropriate? Taking $1,2,\ldots,-2,-1$ will additionally result in some elements being "infinitely far" from the first element. Is this also desirable? – arjafi Jan 29 '13 at 12:52
@CutieKrait: Then you really do not need anything to be well-ordered. Something like what you wrote, $1,2,3,\ldots,-3,-2,-1$, would likely be fine. – arjafi Jan 29 '13 at 13:26