# Is every continuum-sized dense subset of the irrationals order isomorphic to the irrationals?

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?

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.

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.

• 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. Apr 18 at 1:06
• 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. Apr 20 at 1:15
• @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. Apr 20 at 1:20

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.