Pointed rational extension of reals is a homogeneous space

Question

The pointed rational extension of reals is a topology (let us call it PRX topology) on $\mathbb R$ generated by the union of order topologies on $\mathbb Q\cup\{x\}$ for every $x\in\mathbb R$ (cf. pi-Base S60). While it seems natural that rational and irrational numbers are somehow different in this topology, I realised that this space is (apparently) homogeneous (homeomorphisms act transitively on the space). Below is my idea for the proof.

It is enough to find a homeomorphism sending $0$ to $x$ for arbitrarily fixed $x\in \mathbb R$. For $x\in \mathbb Q$ a simple translation by $x$ is a desired homeomorphism, but the construction below also applies. For an irrational $x$, observe that the set $\mathbb Q\cup\{x\}$ is equipped with a dense linear order without extremal elements, hence it is isomorphic to $\mathbb Q$ (Cantor's isomorphism theorem). Let $f:\mathbb Q\to \mathbb Q\cup\{x\}$ be such an order-preserving bijection with $f(0)=x$ (we can fix this value and find respective isomorphisms of the two unbounded open intervals). I suspect that $f$ extends to a homeomorphism of $\mathbb R$ with the PRX topology. It extends to an order preserving bijection of $\mathbb R$ via the same argument as here (take $F(x):=\sup f((-\infty,x)\cap\mathbb Q)$). Then neighbourhoods of $0$ are clearly mapped to neighbourhoods of $x$. For all other points we can consider neighbourhoods enclosed in intervals not containing $0$. Clearly the rationals in such interval are mapped to rationals and the neighbourhoods of the given point are preseved. This argument is valid also for $F^{-1}$, hence $F$ is a homeomorphism.

Is this right? Can this be written in a concise yet more precise way?

Can a similar argument be applied to the pointed irrational extension of reals? pi-Base S61

Answer 1

Let $D$ be either the rationals $\mathbb Q$ or irrationals $\mathbb P$, and let $\mathbb R_D$ have the topology generated by $\{x\}\cup((a,b)\cap D)$ for each $a<x<b$.

Let $p\in\mathbb R_D$. We observe that $D\cup\{p\}=((\leftarrow,p)\cap D)\cup\{p\}\cup((p,\rightarrow)\cap D)$ is order-isomoprhic to $D$. When $D=\mathbb Q$ this is Cantor's isomorhpism theorem, as $D$ and $D\cup\{p\}$ are both infinite countable order-dense sets without a maximum or minimum. When $D=\mathbb P$, we have that $\mathbb R_D\setminus(D\cup\{p\})$ is an infinite countable order-dense set, which is order-isomorphic to $\mathbb Q$; extending this to $\mathbb R$ yields an order-isomorphism from $D\cup\{p\}$ to $\mathbb P=D$.

So let $f_p:D\cup\{p\}\to D$ be an order-isomorphism, which extends to an order-isomorphism $f_p:\mathbb R\to\mathbb R$. This is a bijection; we should verify this is a homeomorphism. For $x\in\mathbb R_D\setminus\{p\}$, let $U=\{x\}\cup((a,b)\cap D))$ be a neighborhood with $(a,b)$ missing $p$, then $f_p[U]=\{f_p(x)\}\cup((f_p(a),f_p(b))\cap D)$ is open. Also, for $U=\{p\}\cup((a,b)\cap D)=\{p\}\cup((a,p)\cap D)\cup((p,b)\cap D)$, we have $f_p[U]=\{f_p(p)\}\cup((f_p(a),f_p(p))\cap D)\cup((f_p(p),f_p(b))\cap D)=\{f_p(p)\}\cup((f_p(a),f_p(b))\cap D)$ open as well. So we have that $f$ is an open map; we may apply similar arguments to show $f^{-1}$ is open as well.

Thus we have shown that for any pair of points $p,q\in\mathbb R_D$, there are homeomorphisms sending $p,q$ to points $f_p(p),f_q(q)\in D$. Finally, for $a,b\in D$, the map $x\mapsto x+b-a$ is a homeomorphism sending $a$ to $b$, completing the proof that $\mathbb R_D$ is homogeneous.