3
$\begingroup$

I was wondering what was the cardinal of the set of the isomorphism classes of the total dense orders on the reals.

I was able to find that it was at least the cardinal of the continuum, because you can take the set $\mathbb{N} \times \mathbb{R}$, which is in bijection with $\mathbb{R}$, and decide for each copy of the reals if you remove all the irrational numbers on the interval $[-1, 1] $ or not. Then, you define the lexicographic order on this final set, with the classical orders on $\mathbb{R}$ and $\mathbb{N}$. An isomorphism preserves the "Dedekind cuts" (the missing irrationals, and the gap between two copies), and keeps them in order, therefore two different sets constructed this way aren’t isormorphic.

I’m also curious if there is a generalized answer for the cardinal of the isomorphism classes for the total dense orders on any cardinal ? I know that the answer for $\aleph_0$ is 4, for the classical orders on $[0, 1]$ ; $]0, 1[$ ; $[0, 1[$ ; $]0, 1]$.

$\endgroup$
2
$\begingroup$

As always, writing the question made me realize how close I was to the answer. In this answer, I will admit the axiom of choice.
Let $\kappa$ be a cardinal larger or equal to $2^{\aleph_0}$. I will show that the cardinal of the isomorphism classes of total dense orders on $\kappa$ is $2^{\kappa}$. Let $C$ be this cardinal. An order on $\kappa$ is a function from $\kappa^2$ to $\{0, 1\}$. Therefore, $C \leq 2^{\kappa^2} = 2^\kappa$. We take the set $\kappa \times \mathbb{R}$. Because $\kappa \geq Card(\mathbb{R})$, there is a bijection between $\kappa$ and $\kappa \times \mathbb{R}$. For each copy of the reals, we can choose to remove all irrational numbers in the interval $[-1, 1]$ or not. This gives us a new set $A$, on which we define a lexicographic order, with the ordinal order of $\kappa$ and the classical order of $\mathbb{R}$ on the subsets of $\mathbb{R}$.

Now, we can look at the partitions of $A$ in two subsets $B$ and $C$, such that all elements in $C$ are greater than the elements in $B$, $B$ has no greatest element, and $C$ has no smallest element. If a such cut occurs inside a copy of the reals, it is because we removed the irrationals in the interval $[-1, 1]$, and so this cut is not isolated : there is a sequence of cuts which converges to it from the left and from the right. If a cut occurs between the copy $k$ of the reals and the copy $k+1$, for $k$ and ordinal, the cut is isolated : there is no sequence which converges to it. Finally, if a cut is just behind the copy $k$ of the reals, with $k$ a limit ordinal, there is a convergent sequence of cuts which converges to it from the left, but not from the right. The cuts between the copies of the reals are well-ordered, by the well order of $\kappa$.

An isomorphism between two sets created that way will map the cuts from one of the set on the cuts of the other. Because the cuts between the copies of the reals are well-ordered, there is an unique isomorphism between the two sets, so the copy of the reals associated with an ordinal $k$ in the first set will be send to the copy of the reals associated with $k$ in the second set. Therefore, the two sets are isomorphic if and only if they are equal.

But that gives us $2^\kappa$ different isomorphism classes. Thus, $C = 2^\kappa$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.