Can an infinite set be transitive, irreflexive, total, and have an upper and lower bound?

Question

I need an infinite structure that can be put into an order with the following properties:

The order must... be transitive, be irreflexive, be total (i.e., every two things share some sort of relation), have an upper bound, have a lower bound, be such that every member with a successor has an immediate successor, be such that every member with a predecessor has an immediate predecessor.

I'm at a loss. I don't know much about infinity, but I would naively think that these conditions force finiteness. However, my instructor told us that it can be proven for polyadic logic that if a finite set of any size can satisfy a schema (say, one that spells out the above conditions), then an infinite set can also satisfy that schema.

Is there a structure that satisfies these conditions? I would think something that looked like this might work: $\lbrace p,p+1,...,q-1,q \rbrace$, where $q$ is the upper bound; $p$ is the lower bound; $\lbrace p,p+1,... \rbrace$ is an infinitely ascending chain; $\lbrace ...,q-1,q \rbrace$ is an infinitely descending chain; and every member of $\lbrace...,q-1,q \rbrace$ is greater than every member of $\lbrace p,p+1,... \rbrace$. I know that an well-ordered series does not contain an infinitely descending chain, but perhaps the descending chain in the example could be isomorphic to a wellordered series with a starting point.

Answer 1

How about $\left\{\frac{1}{n}\ :\ n\ge 3\right\} \cup \left\{1-\frac{1}{n}\ :\ n\ge 3\right\}$? You did not state that the upper or lower bounds had to be in the set.

Answer 2

Let our set be $A\cup B\cup C$, where $A$, $B$, and $C$ are pairwise disjoint sets. Let $A$ be order-isomorphic to the set of non-negative integers with the usual order, $B$ order-isomorphic to the set of integers under the usual order, and $C$ order-isomorphic to the negative integers under the usual order. Declare every element of $A$ less than any element of $B$ or $C$, and every element of $B$ less than every element of $C$.

Answer 3

One straightforward order that satisfies the conditions is $\omega+\omega^*$ (where $\omega^*$ is the standard well-ordering on $\mathbb{N}$ 'backwards'). Equivalently, take $\mathbb{Z}$ and say that every negative number is greater than every positive (or zero) number, and that negative and positive numbers are 'internally' ordered as they are normally. Then $-1$ is the largest element, $0$ is the smallest, and it's relatively easy to see that all of your other conditions are satisfied. This order has a nice mental image of starting at zero, walking off to infinity, 'wrapping around', and walking back in from negative-infinity: $0, 1, 2, 3, \ldots, -3, -2, -1$. Note that the conditions you've given don't force a well-ordering, because they don't include the 'well' portion, that there be no infinite descending chains.

(Postscript: you should be able to show that any well-ordering which satisfies all of your conditions is finite. I'll leave it up to you to figure out why this doesn't actually contradict what your instructor said...)