Do there exist totally ordered sets with the 'distinct order type' property that are not well-ordered?

Solution 1:

There are even densely ordered sets with the DOT property.

Let $\langle X,\le\rangle$ be a linear order, and suppose that there are distinct $x,y\in X$ such that $(\leftarrow,x]\cong(\leftarrow,y]$. Without loss of generality assume that $x<y$, and let $f:(\leftarrow,y]\to(\leftarrow,x]$ be an isomorphism. Clearly

$$h:X\to X:z\mapsto\begin{cases} f(z),&\text{if }z\le y\\ z,&\text{if }y<z \end{cases}$$

is an isomorphism of $X$ into itself. Call a linear order rigid if it admits no non-trivial isomorphism into itself; it follows that every rigid linear order has the DOT property. The following theorem is proved in Brian M. Scott, ‘A characterization of well-orders’, Fundamenta Mathematicæ, Vol. $111$ ($1981$), Nr. $1$, pp. $71$-$76$, freely available here.

Theorem. Let $\kappa\ge\omega$ be such that $2^{<\kappa}=\kappa$. Then there is a rigid dense linear order of power $2^\kappa$.

Such a linear order, being rigid, has the DOT property, but since it’s dense, it is not a well-order.

Of course $2^{<\omega}=\omega$, so there is a rigid dense linear order of power $2^\omega$. With a little work to handle limit cardinals, the generalized continuum hypothesis implies that there is a rigid dense linear order of every uncountable cardinality.

Added: A nice explicit example of a scattered linear order that has the DOT property but is not a well-order isn’t too hard to construct. Let

$$X=\{\langle n,\alpha\rangle:n\in\omega\text{ and }\alpha\in\omega_n\}\;,$$

and define a linear order $\preceq$ on $X$ as follows: for $\langle m,\alpha\rangle,\langle n,\beta\rangle\in X$, $\langle m,\alpha\rangle\preceq\langle n,\beta\rangle$ iff either $m>n$, or $m=n$ and $\alpha\le\beta$.

Suppose that $\langle m,\alpha\rangle\prec\langle n,\beta\rangle$. If $m=n$, it’s easy to see that $\big(\leftarrow,\langle m,\alpha\rangle\big]$ is not isomorphic to $\big(\leftarrow,\langle n,\beta\rangle\big]$. If $m>n$, the interval $\big[\langle m,\alpha\rangle,\langle n,\beta\rangle\big)$ has order type $\omega_m+\beta$, where $\beta<\omega_n<\omega_m$. Suppose that $\langle k,\gamma\rangle\prec\langle m,\alpha\rangle$, and let $I=\big[\langle k,\gamma\rangle,\langle m,\alpha\rangle\big)$. If $k=m$, the order type of $I$ is less than $\omega_m\le\omega_m+\beta$. If $k>m$, then the order type of $I$ is at least $\omega_{m+1}+\alpha>\omega_m+\beta$. Thus, the order type of $I$ cannot be $\omega_m+\beta$, and $\big(\leftarrow,\langle m,\alpha\rangle\big]$ is not isomorphic to $\big(\leftarrow,\langle n,\beta\rangle\big]$.

Thus, $\langle X,\preceq\rangle$ has the DOT property. A rough sketch of $X$:

$$\cdots\underset{\omega_3}\longrightarrow\underset{\omega_2}\longrightarrow\underset{\omega_1}\longrightarrow\underset{\omega}\longrightarrow$$