Continuum family of continuum subsets of $\mathbb R$ which are not pairwise order isomorphic

Question

I need to construct a continuum family of pairwise order inequivalent subsets of $\mathbb R$, such that cardinality of intersection of every constructed subset and every nontrivial interval is also continuum.

I was able to construct a countable family of such subsets: for any natural $n$ let $B \subset \mathbb R$ be obtained by throwing away $n$ points. For example, for $n = 1$ we can take $B = \mathbb R \setminus \{0 \}$, and for $n = 2$ $B' = \mathbb R \setminus \{0, 1\}$. Then $B$ has a subset without $sup$, one subset without $inf$, and $B'$, in addition, has a subset without both $inf$ and $sup$.

I think we should construct similar sets which differ in second order properties, but I failed to do so.

Thanks!

Answer 1

The family can be constructed by recursion by killing off each potential order-isomorphism.

Suppose that $A$ and $B$ are dense subsets of $\Bbb R$, and $f:A\to B$ is an order-isomorphism. Since $A$ and $B$ are dense in $\Bbb R$, their order topologies coincide with their subspace topologies. Define

$$\hat f:\Bbb R\to\Bbb R:x\mapsto\sup\{f(y):y\in A\cap(\leftarrow,x]\}\,.$$

Clearly $\hat f$ is monotone, so any discontinuity would have to be a jump discontinuity, and the fact that $B$ is dense in $\Bbb R$ makes that impossible, so $\hat f$ must be continuous.

Each continuous function from $\Bbb R$ to $\Bbb R$ is determined by its restriction to $\Bbb Q$, and there are $|\Bbb R|^{|\Bbb Q|}=\left(2^{\omega}\right)^{\omega}=2^\omega$ functions from $\Bbb Q$ to $\Bbb R$, so there are just $2^\omega$ continuous functions from $\Bbb R$ to $\Bbb R$; let $F=\{f_\xi:\xi<2^\omega\}$ be an enumeration of them. Let $\mathscr{I}=\{I_n:n\in\omega\}$ be an enumeration of the open intervals in $\Bbb R$. Finally, let $\{\langle\alpha_\xi,n_\xi,\beta_\xi,\gamma_\xi\rangle:\xi<2^\omega\}$ be an enumeration of $2^\omega\times\omega\times 2^\omega\times 2^\omega$.

Recursively choose points $x_\xi\in\Bbb R$ for $\xi<2^\omega$ as follows. If $\eta<2^\omega$, and distinct points $x_\xi$ have been chosen for $\xi<\eta$, let $X_\eta=\{x_\xi:\xi<\eta\}$, let

$$Y_\eta=X_\eta\cup\bigcup_{\xi\le\eta}f_{\alpha_\xi}[X_\eta]\cup\bigcup_{\xi\le\eta}f_{\alpha_\xi}^{-1}[X_\eta]\,,$$

and let $x_\eta\in I_{n_\eta}\setminus Y_\eta$. (This is possible, since $|Y_\eta|<2^\omega=|I_{n_\eta}|$.) Clearly this construction goes through to $2^\omega$.

For each $\eta<2^\omega$ let $A_\eta=\{x_\xi:\beta_\xi=\eta\}$. For each $\eta<2^\omega$ and $n\in\omega$ we have $x_\xi\in A_\eta\cap I_n$ for each $\xi<2^\omega$ such that $n_\xi=n$ and $\beta_\xi=\eta$, so $|A_\eta\cap I_n|=2^\omega$. That is, each of the sets $A_\eta$ meets each non-empty open interval in a set of cardinality $2^\omega$.

Now suppose that $f:A_\mu\to A_\nu$ is an order isomorphism for some $\mu,\nu<2^\omega$; there is an $\eta<2^\omega$ such that $f_{\alpha_\eta}=\hat f$, and there is a $\xi<2^\omega$ such that $x_\xi\in A_\mu$ and $\eta\le\xi$. Let $\zeta<2^\omega$; if $\zeta<\xi$, then $x_\zeta\in X_\xi$, so $x_\xi\ne f_{\alpha_\eta}^{-1}(x_\zeta)$, and hence $f(x_\xi)\ne x_\zeta$. And if $\xi<\zeta$, then $x_\xi\in X_\zeta$, so $x_\zeta\ne f_{\alpha_\eta}(x_\xi)=f(x_\xi)$. Thus, $f(x_\xi)=x_\xi$, and hence $\mu=\nu$, and it follows that $\{A_\eta:\eta<2^\omega\}$ is a family of pairwise non-isomorphic orders.

Answer 2

Think in terms of "coding." Let's say we just wanted continuum-many non-order-isomorphic sets of reals. One thing we could do is the following. Given an infinite binary sequence $\alpha=(a_i)_{i\in\mathbb{N}}$, consider a set of reals $X_\alpha$ whose ordertype is of the form $$\mathbb{Q}+{\bf (a_1+2)}+\mathbb{Q}+{\bf (a_2+2)}+\mathbb{Q}+{\bf (a_3+2)}+\mathbb{Q}+...$$ where ${\bf n}$ denotes a linear order with $n$ elements (the "$+2$" is needed since otherwise we "lose" blocks - think about how $\mathbb{Q}+{\bf 1}+\mathbb{Q}\cong\mathbb{Q}$).

It's not hard to show that we can "read back" $\alpha$ from $X_\alpha$'s isomorphism type; consequently, if $\alpha\not=\beta$ then $X_\alpha\not\cong X_\beta$ and so $\{X_\alpha: \alpha\in 2^\mathbb{N}\}$ does the job.

Now of course this won't solve your problem, since the "size condition" isn't satisfied. But can you see another way to "code" a sequence of numbers into a linear order which will do the job here?

HINT: let $I\subseteq\mathbb{R}$ be a set of reals such that for every interval $(a,b)$ we have $\vert I\cap (a,b)\vert=\vert (a,b)\setminus I\vert=2^{\aleph_0}$. Now any linear order containing $I$ will satisfy the relevant size condition since $I$ is "big enough," and you have room to do some coding since the complement of $I$ is "big enough."