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Let $a,b\in\mathbb{R}$. Is $(a,b)\setminus \mathbb{Q}$ homeomorphic to $\mathbb{R}\setminus \mathbb{Q}$?

It seems true to me from this fact: $(a,b)$ is homeomorphic to $\mathbb{R}$. Let $f$ be this homeomorpshim then $f|_{(a,b)\setminus\mathbb{Q}}$ is a homeomorphsim from $(a,b)\setminus\mathbb{Q}$ to it's image under $f$. Of course, the image does have to be made up of just irrational, but is that possible?

UserA
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  • I've marked this as a duplicate, because most of the given answers there are in fact homoemorphisms. – Asaf Karagila Apr 16 '18 at 09:04
  • There are also the topological characterizations of the irrationals, which then provide you with a way of proving the two spaces are homeomorphic. – Asaf Karagila Apr 16 '18 at 09:05
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    The linked question only deals with rational endpoints $a$ and $b$. It's true even when one or both endpoints are irrational. – Angina Seng Apr 16 '18 at 09:06

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The set $(a,b)\cap \Bbb Q$ is order-isomorphic to $\Bbb Q$, since both are countable, densely ordered sets with no endpoints. Pick an order isomorphism $F:(a,b)\cap \Bbb Q\to\Bbb Q$. This extends uniquely to an order isomorphism $G:(a,b)\to\Bbb R$: for $x\in(a,b)$ define $G(x)=\sup\{F(t):t\in(a,b)\cap\Bbb Q,t\le x\}$. As $(a,b)$ and $\Bbb R$ have the order topology, then $G$ is a homeomorphism. Its restriction to $(a,b)\setminus\Bbb Q$ is a homeomorphism to $\Bbb R\setminus\Bbb Q$.

Angina Seng
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  • The result "since both are..." in your first sentence is a classic from Georg Cantor. The first step in one elementary proof is to show that a countably infinite ordered set $S$ without endpoints has a subset, unbounded above & below, that is order-isomorphic to $\Bbb Z.$ This step can be done in ZF if $S\subset \Bbb R$ with the usual order. I have been wondering whether this step can be done in ZF in the general case, and (equivalently) wondering whether Cantor's result is provable in ZF. – DanielWainfleet Apr 16 '18 at 18:56
  • I just figured out an answer to my query in my comment, It $is$ provable in ZF. Let $S={s_n: n\in \Bbb N}.$ Let $f(1)=1.$ Let $f(n+1)$ be the least $m$ such that $s_m>s_{f(n)}.$ Then ${s_{f(n)}:n\in \Bbb N}$ is unbounded above and order-isomorphic to $\Bbb N.$ – DanielWainfleet Apr 16 '18 at 19:08