Is every subset of $\mathbb{Q}$ order-isomorphic to a closed subset of $\mathbb{Q}$?

Question

This question was inspired by a question about countable subsets of the real line, in which the countable sets were compared to countable linear orders. It occurred to me that every such subset I could think of appeared to be order-isomorphic to a subset of $\mathbb{Q}$ that is closed in $\mathbb{Q}$.

I've come up with a rough idea for a proof, but I'm not certain it will actually work. I'm curious if there is an easy/standard proof or counterexample to the question:

Is every subset $S \subset \mathbb{Q}$ order-isomorphic to a subset $C \subset \mathbb{Q}$ such that $C$ is closed as a subspace of $\mathbb{Q}$?

I'll share my rough idea for a proof: We identify points in $\bar{S} - S$ in $\mathbb{Q}$ as limit points of $S$ in $\mathbb{R}$, which lie in $\mathbb{Q}$ but not $S$. Clearly $S$ must be countable, so we can enumerate these points, and we attempt to "remove" them one by one, by performing order-preserving operations that "move these points" to irrational numbers.

For each such point we translate $S$ in a possibly-piecewise fashion so that the sequences approaching this point of discontinuity approach an irrational number instead, then doing some kind of additional operations to get the rest of the points back in $\mathbb{Q}$. But it's not clear if we can do this in a way that gives us a "limit function" that is well-defined on $S$, nor if this approach can be done in a way that's "stable" enough that we can remove each bad point without creating new ones.

Edit: Wait a minute, instead of worrying about irrational numbers, can I just replace the $n^{\text{th}}$ element of $\bar{S} - S$ with an interval of size $\frac{1}{2^n}$ by adding $\frac{1}{2^n}$ to everything bigger than it? Addendum to edit: No I can't, two reasons why are provided in the comments. I'm not sure what I was thinking there.

Answer 1

One way is the following. You have that $(\mathbb{Q}\times\mathbb{Q},<_{lex})$ is order isomorphic to $\mathbb{Q}$ with the standard order by Cantor's theorem. Given $S\subseteq\mathbb{Q}$ then we can embed it into $\mathbb{Q}\times\mathbb{Q}$ via the map $s\mapsto (s,0)$. We have that $S\times\{0\}$ will be closed in $\mathbb{Q}\times \mathbb{Q}$, since any $(a,b)\notin S\times \{0\}$ either $a\notin S$ and $a\times\mathbb{Q}$ will be an open neighborhood of $(a,b)$ disjoint from $S\times\{0\}$, or $b\neq 0$ and $((a,b-\epsilon),(a,b+\epsilon))$ will be an open neighborhood of $(a,b)$ disjoint from $S\times\{0\}$. Of course $S$ and $S\times\{0\}$ are order isomorphic. Let $f:\mathbb{Q}\times\mathbb{Q}\rightarrow \mathbb{Q}$ be an order isomorphism, it will also be a homeomorphism. So $f(S\times\{0\})$ will be the desired set.

Edit: I will add a quick sketch as to why order isomorphisms are homeomorphisms. Let $L$ and $P$ be linear orders and $f:L\rightarrow P$ be an order isomorphims. For all $p,q\in P$ we such that $p<_P q$ we have that $(f^{-1}(p),f^{-1}(q))=f^{-1}((p,q)_{<_P})$. That is, the preimage of of a basic open set of $P$ is a basic open set of $L$, so $f$ is continuous. Similarly $a,b\in L$ such that $a<_L b$ we have that $f((a,b))_{<_L}=(f(a),f(b))_{<_P}$. That is, we have that the image of a basic open set of $L$ is a basic open set of $P$, so $f$ is open. $f$ is bijective since it is an order isomorphism, so $f$ will be a homeomorphism.