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Cross-posting this question from MSE.

Let $C\subseteq \mathbb R$ denote the Cantor set and let $A\subseteq C$ be dense with $|A| = |\mathbb R|$, and assume $A$ doesn't contain its infimum. Is there a locally compact, second countable, Hausdorff topology on $A$ that is finer than the right order topology (the topology with base $\{(a,\infty)\cap A: a\in A\}$)? If it helps, the continuum hypothesis can be assumed.

An equivalent formulation of the problem is to find a compact metric on $A\cup \{0\}$ refining the right order topology by identifying the point 0 with the point added in the 1-point compactification.

If $|C\setminus A|$ is finite, the result is easy as $C$ is compact in the Euclidean topology. If $C\setminus A$ is countable, then the same methods as this answer should hold. The difficulty is if $|C\setminus A| = |\mathbb R|$.

This question is a special case of this question.

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This is not always possible (even under extra assumptions such as the continuum hypothesis). In particular, $\def\ZFC{\mathsf{ZFC}}\ZFC$ proves the following:

Proposition. If $A \subseteq C$ is a Bernstein set, then there is no complete metric on $A$ refining the right order topology.

Proof. We'll prove a stronger contrapositive: If there is a complete metric on an uncountable set $A$ refining the right order topology, then there is an uncountable set $F \subseteq A$ that is closed in the Euclidean topology.

Assume that $d$ is a complete metric on $A$. Since $d$ refines the right order topology, we have that the inclusion map $\iota : (A,d) \to (C,|\cdot|)$ is lower semicontinuous and so in particular Borel-measurable. This implies that $\iota$ as a subset of $(A,d) \times (C,|\cdot|)$ is Borel. Since $A$ is uncountable and since Borel sets have the perfect set property, there is a perfect set $G \subseteq \iota$. Let $F$ be the projection of $G$ to $C$. Note that since $\iota$ is a literal inclusion map, we just have that $G = \{(x,x) : x \in F\}$. In particular $F$ is an uncountable subset of $A$ that is closed (in both topologies).

Since Bernstein sets have no uncountable closed subsets, this implies that no such $d$ can exist if $A$ is a Bernstein set. $\square$

Finally, note that $\ZFC$ proves the existence of Bernstein set with the same cardinality as $\mathbb{R}$ without any additional set-theoretic assumptions.

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  • $\begingroup$ Why must $F$ be closed in $C$? Projections of closed sets need not be closed. If you consider the embedding map $\iota:\mathbb Q^c\to \mathbb R$ (Euclidean on domain and codomain), then $\{(x,x):x\in \mathbb Q^c\cap[0,1]\}\subseteq \iota$ is closed in $\mathbb Q^c\times \mathbb R$, but $\mathbb Q^c\cap[0,1]$ is not closed in $\mathbb R$. It is closed in the subspace topology, but your argument seems to require $F$ to be closed in the topology on $C$, not on the subspace topology on $A$. $\endgroup$ Commented May 12 at 1:23
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    $\begingroup$ @daRoyalCacti $G$ can be taken to be compact, which is enough to ensure that it's closed in both topologies. $\endgroup$ Commented May 12 at 2:06
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    $\begingroup$ Compact will do it. Thank you $\endgroup$ Commented May 12 at 2:10

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