Post History
#2: Post edited
by
clemens
·
2026-03-11T05:03:42Z (6 months ago)
- I think the question is probably clear enough from the title, so here I'll just give some of the context that prompted me to ask it and then write what I've been able to find so far.
- [A relatively recent question](https://math.codidact.com/posts/293500) asked whether $\mathbb{R}$ has a topological field structure other than the usual one. The answer to the question as stated is obviously "Yes", but I and other answerers tried to figure out the more interesting question of *what* the different topological field structures on $\mathbb{R}$ could be.
- Unfortunately, [my answer](https://math.codidact.com/posts/293500/295549#answer-295549) was incomplete, in particular w.r.t. its "starting-point": how we can topologize $\mathbb{Q}$. Hence I'm asking this question to give more attention to what I hope will be an interesting and enjoyable subproblem of the original problem posed a year ago.
- I currently have a "upper" and a "lower" bound, so to speak, for this problem.
- 1. The "upper" bound is provided by Urysohn's metrization theorem: since the reals are second-countable, they are metrizable. Hence any topological field structure on $\mathbb{Q}$ must be a metric topology.
2. The "lower" bound is provided by the various completions of $\mathbb{Q}$ (the Archimedean completion $\mathbb{R}$ with the usual topology thereon and the $p$-adic completions $\mathbb{Q}_p$ again with the usual topology) and by finite or infinite products thereof. Considering $\mathbb{Q}$ as a subspace either of any of these complete topological fields or of their products gives rise to a unique topological field structure on $\mathbb{Q}$.- Thus, the challenge is either to somehow refine this "upper bound" (e.g. by showing that the metric given by any topological field on $\mathbb{Q}$ induces a norm, not trivially as the Kuratowski embedding would) or increase this "lower bound" (by providing examples of topologies that cannot be induced by these obvious embeddings of $\mathbb{Q}$).
- I think the question is probably clear enough from the title, so here I'll just give some of the context that prompted me to ask it and then write what I've been able to find so far.
- [A relatively recent question](https://math.codidact.com/posts/293500) asked whether $\mathbb{R}$ has a topological field structure other than the usual one. The answer to the question as stated is obviously "Yes", but I and other answerers tried to figure out the more interesting question of *what* the different topological field structures on $\mathbb{R}$ could be.
- Unfortunately, [my answer](https://math.codidact.com/posts/293500/295549#answer-295549) was incomplete, in particular w.r.t. its "starting-point": how we can topologize $\mathbb{Q}$. Hence I'm asking this question to give more attention to what I hope will be an interesting and enjoyable subproblem of the original problem posed a year ago.
- I currently have a "upper" and a "lower" bound, so to speak, for this problem.
- 1. The "upper" bound is provided by Urysohn's metrization theorem: since the reals are second-countable, they are metrizable. Hence any topological field structure on $\mathbb{Q}$ must be a metric topology.
- 2. The "lower" bound is provided by the various completions of $\mathbb{Q}$ (the Archimedean completion $\mathbb{R}$ with the usual topology thereon and the $p$-adic completions $\mathbb{Q}_p$ again with the usual topology) and by finite or infinite products thereof. Different embeddings of $\mathbb{Q}$ (as a topological ring) into a product of such topologies will induce different topological field structures on $\mathbb{Q}$.
- Thus, the challenge is either to somehow refine this "upper bound" (e.g. by showing that the metric given by any topological field on $\mathbb{Q}$ induces a norm, not trivially as the Kuratowski embedding would) or increase this "lower bound" (by providing examples of topologies that cannot be induced by these obvious embeddings of $\mathbb{Q}$).
#1: Initial revision
by
clemens
·
2026-03-11T05:00:13Z (6 months ago)
Topological field structures on $\mathbb{Q}$ not generated by the $p$-adic and Archimedean topologies?
I think the question is probably clear enough from the title, so here I'll just give some of the context that prompted me to ask it and then write what I've been able to find so far.
[A relatively recent question](https://math.codidact.com/posts/293500) asked whether $\mathbb{R}$ has a topological field structure other than the usual one. The answer to the question as stated is obviously "Yes", but I and other answerers tried to figure out the more interesting question of *what* the different topological field structures on $\mathbb{R}$ could be.
Unfortunately, [my answer](https://math.codidact.com/posts/293500/295549#answer-295549) was incomplete, in particular w.r.t. its "starting-point": how we can topologize $\mathbb{Q}$. Hence I'm asking this question to give more attention to what I hope will be an interesting and enjoyable subproblem of the original problem posed a year ago.
I currently have a "upper" and a "lower" bound, so to speak, for this problem.
1. The "upper" bound is provided by Urysohn's metrization theorem: since the reals are second-countable, they are metrizable. Hence any topological field structure on $\mathbb{Q}$ must be a metric topology.
2. The "lower" bound is provided by the various completions of $\mathbb{Q}$ (the Archimedean completion $\mathbb{R}$ with the usual topology thereon and the $p$-adic completions $\mathbb{Q}_p$ again with the usual topology) and by finite or infinite products thereof. Considering $\mathbb{Q}$ as a subspace either of any of these complete topological fields or of their products gives rise to a unique topological field structure on $\mathbb{Q}$.
Thus, the challenge is either to somehow refine this "upper bound" (e.g. by showing that the metric given by any topological field on $\mathbb{Q}$ induces a norm, not trivially as the Kuratowski embedding would) or increase this "lower bound" (by providing examples of topologies that cannot be induced by these obvious embeddings of $\mathbb{Q}$).