Compact metric topology on irrationals

Question

Is there a compact metric topology on $I=(\mathbb Q^c \cap (0,\infty))\cup \{0\}$ which is finer than the topology of half rays: $\{(a, \infty)\cap \mathbb Q^c: a\in I\}\cup\{\emptyset, I\}$?

In the topology of half rays, I is compact as I is the only open set containing 0. However, this topology isn't Hausdorff and so in particular isn't metrisable.

The discrete topology is metrizable and is finer than the half-ray topology, but it isn't compact. Similarly, for the Euclidean topology. The topology induced by bijecting $I$ into $[0,1]\subseteq\mathbb R$ is compact and metrisable but (unlikely) isn't finer than than the half ray topology. It seems like a purpose-built topology is needed, but I can't come up with one.

An idea I haven't had much success with is to find a locally compact, second countable, Hausdorff topology on $\mathbb Q^c\cap(0,\infty)$ and then identifying 0 with the point added in the one point compactification. I believe this is an equivalent formulation of the problem, as the compactification will be second countable, compact, Hausdorff and hence metric. Using the discrete topology for this fails as it isn't second countable, and using the Euclidean topology also fails as local compactness fails.

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Update: New things that I've tried

1. Decomposing $I$ into countably many sets, each order isomorphic to $\mathbb R$ (Decomposition of irrationals into sets order isormorphic to $\mathbb R$), then inducing a topology by each of these order isomorphisms. Seems to fail by the Baire category theorem.
2. Trying to generate a continuous bijection (Bijection from irrationals to reals mapping rays to open sets) and then inducing a topology with the bijection. Haven't made any progress with this idea. Update: this method seems promising based on the solution by Alex Ravsky to the other question.
3. Some useful properties of the right-order topology that I haven't found a use for: $I$ is hyper-connect, ultra-connected, path-connected, locally connected, pseudo compact, locally compact, $T_0, T_4, T_5$ (using the definition of these in Steen's counter-examples in topology), not $T_1,T_2,T_3$.
4. The right order topology is induced by a quasi-pseudo-metric: $q(x,y) = x-y$ if $x\geq y$ and $q(x,y) = 0$ otherwise. Might be something in metric refinements of quasi-pseudo-metric spaces. I have yet to find any useful theory on this.

Posted a more general question to MathOverflow (https://mathoverflow.net/questions/491619/locally-compact-second-countable-hausdorff-topology-refining-the-right-order-t)

Answer 1

We can construct the required topology following your idea 2. The map $\exp:\mathbb R\to (0,\infty)$, $x\mapsto e^x$ is a homeomorphism, so the set $Q$, $Q=\{\ln q:q\in\mathbb Q\cap (0,\infty)\}$ is dense in $\mathbb R$. By Proposition from the answer to your other question, there exists a bijection $f:\mathbb R\to\mathbb R\setminus Q$ such that for each $a\in\mathbb R$ the set $f^{-1}((a,\infty))$ is open. Let $g:\mathbb R\to I\setminus \{0\}$ be the composition $\exp\circ f$. Then $g$ is a bijection such that for each $a\ge 0$ the set $g^{-1}((a,\infty))$ is open. Let $\tau$ be a unique topology on $I\setminus \{0\}$ such that $g$ is a homeomorphism. Then $(a, \infty)\cap I\in\tau$ for each $a\ge 0$. Identifying $0$ with the added point of the Alexandrov compactification of $(I\setminus\{0\},\tau)$ we obtain a space $I$ homeomorphic to the Alexandrov compactification of the reals, which is a metric compact.