Anti-dense linear orders

Question

Suppose $(L,\leq)$ is a linear order that is anti-dense, meaning, between any two distinct points $x$ and $y$ there are only finitely many elements in between those two elements. Then, is it true that either $(L,\leq)$ is a finite linear order, or it is isomorphic to the positive integers under the usual linear order, or it is isomorphic to the negative integers under the usual linear order, or it is isomorphic to the integers?

Answer 1

Yes.

Let $x \in L$. Then either $x$ is the greatest element in $L$, or there is some $z > x$. As in the second case there are only finitely many elements $y \in L$ such that $x < y \leqslant z$, we can pick the least of them and name it $S(x)$. This is the successor of $x$, i.e. the least element greater than $x$.

Similarly, every element $x \in L$ except the least one has a predecessor $P(x)$.

Also for $x \in L$ denote $P^k(x) = \underbrace{P(P(\ldots P(x) \ldots))}_{k \text{ times}}$ for $k \in \mathbb{N}$, similarly $S^k(x)$.

Now

• If there exist the least element $a \in L$ and the greatest element $b \in L$, there is finitely many $x \in L$ such that $a < x < b$, hence $L$ is finite.

• If there is the least element $a \in L$ but no greatest element, $S(x)$ is defined for all $x \in L$. Consider the set $L_0 = \{ S^k(a) : k \in \mathbb{N} \}$. One easily shows that $L_0 = L$ and the function $f : \mathbb{N} \to L_0$ given by $f(k) = S^k(a)$ is an isomorphism, so $L \cong \mathbb{N}$.

• A similar proof applies to the case when there is a greatest element in $L$ but no least element, yielding that $L$ is isomorphic to the negative integers.

• Suppose there is neither a greatest element in $L$ nor the least. Fix an arbitrary $a \in L$ and consider $L_0 = \{ P^k(a) : k \in \mathbb{N} \} \cup \{ S^k(a) : k \in \mathbb{N} \}$. Then again, one easily shows that $L_0 = L$ hence $L \cong \mathbb{Z}$.