# Are uncountable dense subsets of $\mathbb{R}$ with the same cardinality homeomorphic?

I know that if $A,B\subseteq\mathbb{R}$ are two dense countable subsets of reals they are homeomorphic, indeed there exists a homeomorphism $f:\mathbb{R}\to\mathbb{R}$ such that $f(A)=B$. I was wondering what happens for cardinalities different from $\aleph_0$, i.e. if I have two dense sets $A,B\subseteq\mathbb{R}$ of the same cardinality $\kappa$, then is true that there exists a homeomorphism $f:\mathbb{R}\to\mathbb{R}$ such that $f(A)=B$?

To avoid trivialities, $\kappa$ must be infinite because the sets are dense; and clearly a necessary condition is that holds $\kappa<\mathfrak{c}$, because otherwise $A=\mathbb{R}\setminus\{0\},B=\mathbb{R}\setminus\mathbb{Q}$ form a cunterexample.

Then if the continuum hypotesis is assumed, we already know the answer; but without it the question is interesting and I don't know the answer.

• What property will $f$ satisfy? $f$ is a homeomorphism? – Idonknow Jun 2 '18 at 15:58
• Yes, $f$ homeomorphism, i forgot it. – G. Ottaviano Jun 2 '18 at 16:43
• I've improved the title. "A question about ..." might not attract much attention. – symplectomorphic Jun 2 '18 at 16:47
• It was a good idea, i didn't know what to call it, that is a very much better title, thanks. – G. Ottaviano Jun 2 '18 at 17:06

This is a very interesting question and the subject of current research in set theory.

There are, however, some caveats.

Say that a set of reals is $\aleph_1$-dense if and only if it meets each interval in exactly $\aleph_1$-many points. It is easy to see that such sets exist, have size $\aleph_1$, and in fact, if $A$ is $\aleph_1$-dense, then between any two points of $A$ there are precisely $\aleph_1$-many other points of $A$. Now, if $A$ is $\aleph_1$-dense in $(-\infty,0)$, $B$ is countable dense in $[0,1]$, $C$ is $\aleph_1$-dense in $(1,\infty)$, and $D$ is $\aleph_1$-dense in $\mathbb R$, then $|A\cup B\cup C|=|D|=\aleph_1$ but they are not homeomorphic.

This suggests that the right generalization is not quite what you suggest, but rather than asking about dense sets of size $\kappa$, we should instead ask about $\kappa$-dense sets (the obvious generalization of the notion of $\aleph_1$-dense from the previous paragraph).

Shelah proved in the early 1980s that if $2^{\aleph_0}<2^{\aleph_1}$ then there is a family of $2^{\aleph_1}$ mutually incomparable $\aleph_1$-dense sets (meaning that none of them embeds in an order-preserving fashion in any of the others). Since the assumption is consistent with the failure of CH, this shows that it is consistent with the standard axioms of set theory that the property you are asking about fails badly. The result can be strengthened to ensure that there is such a family of mutually incompatible $\aleph_1$-dense sets, where incompatibility means that for no pair of orders in the family there is no uncountable ordered set that embeds in both.

On the other hand, in 1973 Baumgartner proved that it is consistent that all $\aleph_1$-dense sets of reals are order-isomorphic. This was later strengthened to show the consistency of this statement with Martin's axiom (and with Martin's axiom and $2^{\aleph_0}>\aleph_2$) and even with the proper forcing axiom.

The version in the question looks stronger: we ask not just that the sets be isomorphic but in fact that there is an order automorphism of $\mathbb R$ that witnesses this. Actually, this strengthening follows from the apparently weaker version because (one can easily check that) any order-preserving isomorphism between dense sets of reals extends (in a unique fashion) to an order-automorphism (necessarily, an auto-homeomorphism) of $\mathbb R$. Thanks to Ashutosh Kumar for setting me straight.

The above addresses the first case of your question, $\kappa=\aleph_1$. It turns out this is really all we know so far. There is active work on the case $\kappa=\aleph_2$ (again, in the version asking whether it is consistent that any two $\aleph_2$-dense sets of reals are order-isomorphic). Moore and Todorcevic have a recent paper on the subject, and Itay Neeman has (tentatively) announced its consistency, using his technique of forcing with models of various sizes as side conditions, but as far as I know there is no paper yet.

• What does it means mutually incomparable? Incomparable by inclusion? – G. Ottaviano Jun 2 '18 at 17:04
• By embeddability: neither one injects in an order-preserving fashion in the other. – Andrés E. Caicedo Jun 2 '18 at 17:15
• This addresses your remark "I expect that the stronger version ... is false". Every order preserving bijection between two dense sets of reals has a unique extension to an order preserving bijection of $\mathbb{R}$. – hot_queen Jun 6 '18 at 10:33
• @hot Ah, you are right. It is actually obvious, I don't know why I was worried. Sigh Thanks. I'll edit the answer in a little bit. – Andrés E. Caicedo Jun 6 '18 at 11:08