Noninital ordinal of a set

Question

If $\mathbb R$ is well ordered by an order $\ll$ such that the ordinal of R and that order $\alpha$ is not initial. How can I show that there is a closed interval $[a,b]$ ($a<b$) such that the ordinal of that interval with the order $\ll$ isn't $\alpha$? If $\alpha$ is inital ordinal I can easily show that for all such intervals their ordinal with $\ll$ is $\alpha$ but I cannot find counter example with explicit order of $\mathbb R$ to prove my claim.

Answer 1

Assume the continuum hypothesis, for simplicity. Let $A\subseteq\Bbb R$ be such that both $A$ and $\Bbb R\setminus A$ meet every open interval on an uncountable set (for example, if $A$ is a Bernstein set).

We order $\Bbb R$ in the order type $\omega_1+\omega_1$. Take some bijection $f$ of $\Bbb R$ with $\omega_1$ then apply it twice, to $A$ and to its complement, decreeing that $x\ll y$ if $x\in A\land y\notin A$, or if both in the same part and $f(x)<f(y)$.

Easily $(\Bbb R,\ll)$ is of order type $\omega_1+\omega_1$. Now suppose that $[a,b]\subseteq\Bbb R$ is a non-trivial interval, then $[a,b]\cap A$ is uncountable, and therefore has order type $\omega_1$ and $[a,b]\setminus A$ is uncountable and therefore must have order type $\omega_1$. This is because every uncountable subset of $\omega_1$ has order type $\omega_1$.

It follows that $[a,b]$ with the restriction of $\ll$ is also of order type $\omega_1+\omega_1$.

Answer 2

Are you sure that what you're trying to prove is true? It looks false to me, for the following reason. Let $\mathfrak c$ denote the cardinal of the continuum. It is easy (by transfinite recursion) to partition $\mathbb R$ into two sets $A$ and $B$, each of which has $\mathfrak c$ points in every nondegenerate interval. Now well-order $\mathbb R$ as follows. First well-order each of $A$ and $B$ with order-type $\mathfrak c$ (initial ordinal); then put all the elements of $B$ after all the elements of $A$. So you get a well-ordering of $\mathbb R$ with order-type $\mathfrak c+\mathfrak c$ (non-initial). Furthermore, every non-degenerate interval will have this same order-type.