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This is a strengthening of a question another user asked, here: Are irrational numbers order-isomorphic to real transcendental numbers?. In the answer to that question, it was stated that the irrationals are order-isomorphic to the transcendental reals. My question is this. Suppose $S$ is a continuum-sized subset of the set of irrational numbers, which has the property that it is everywhere dense, meaning, between any two distinct reals, there exists a real number belonging to $S$. Must $S$ be order-isomorphic to the irrationals?

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Here's a counterexample. Start with the irrationals. Then remove all irrational numbers between $0$ and $1$. Now put back a subset of the irrationals between $0$ and $1$ that is dense and countable. The resulting set satisfies all of your hypotheses, but it has a countably infinite subinterval. No subinterval of the irrationals is countably infinite.

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Building on Lee’s answer, I’ll show that this happens iff $S$ is co-countable. In fact, building on the answer from the linked question, we have the following:

Lemma: If $I_1, I_2$ are two order-dense subsets of $\mathbb{R}$, then any order-isomorphism $f: I_1 \to I_2$ extends to an order-isomorphism $g$ from $\mathbb{R}$ to itself. In particular, $\mathbb{R}\setminus I_1$ and $\mathbb{R}\setminus I_2$ are order-isomorphic.

Proof: Define $g(x) = \sup \{f(w): w < x, w \in I_1\}$. One may check that this is indeed an order-isomorphism extending $f$. $\square$

Thus, if $S$ is an order-dense subset of $\mathbb{R}$ that’s order-isomorphic to the irrationals, then $\mathbb{R}\setminus S$ is isomorphic to $\mathbb{Q}$. In particular, $S$ is co-countable. Conversely, if $S \subset \mathbb{R} \setminus \mathbb{Q}$ is co-countable, then $\mathbb{R} \setminus S$ is countable and contains $\mathbb{Q}$ (therefore order-dense in $\mathbb{R}$). All countable order-dense subsets of $\mathbb{R}$ are order-isomorphic (since they are both countable linearly-ordered sets that are dense in themselves and have no endpoints), so $\mathbb{R} \setminus S$ and $\mathbb{R} \setminus \mathbb{Q}$ are order-isomorphic. Applying the lemma again, we see that $S$ is order-isomorphic to the irrationals.

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  • $\begingroup$ A corollary of this proof is that subsets of $\mathbb{R}$ that are order-isomorphic to the irrationals are exactly those whose complements are countable and order-dense. $\endgroup$
    – David Gao
    Apr 18 at 1:06
  • $\begingroup$ I don't think that is quite true. For example, the set of irrationals between $0$ and $1$ is isomorphic to the entire set of irrationals, but its complement is uncountable, and also it is not everywhere dense. $\endgroup$
    – user107952
    Apr 20 at 1:15
  • $\begingroup$ @user107952 Ah, yes, you’re absolutely right. The proof in my answer is fine, but I over-generalized in my comment. The correct statement should have been that dense subsets of $\mathbb{R}$ that are order-isomorphic to the irrationals are exactly those whose complements are countable and order-dense. $\endgroup$
    – David Gao
    Apr 20 at 1:20
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Let $P$ be a nonempty perfect set of irrational numbers. Then the set $S=P+\mathbb Q=\{p+q:p\in P,\ q\in\mathbb Q\}$ is a set of irrational numbers whose intersection with every interval has the cardinality of the continuum. Since $S$ is $\sigma$-compact, it is neither homeomorphic nor order-isomorphic to the irrational numbers.

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