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Q&A Are there other topologies on $\mathbb R$ that make it a topological field?

posted 7mo ago by clemens‭ · edited 6mo ago by clemens‭

Answer

#6: Post edited by $user avatar$ clemens‭ · 2026-03-08T21:06:51Z (6 months ago)

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Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
Since $\mathbb{Q}$ is second-countable and $T_3$, it is metrizable by Urysohn's theorem, for a proof of which see the nice exposition at the [nLab](https://ncatlab.org/nlab/show/Urysohn+metrization+theorem) ([archived copy](http://web.archive.org/web/2020/https://ncatlab.org/nlab/show/Urysohn+metrization+theorem)).
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a topological field has to follow either the usual topology or some $p$-adic topology, unfortunately requires the metric to be multiplicative, and I don't see how to show that given the kinds of metrics we get from metrization theorems. I don't even see how to show that we can necessarily get a norm on $\mathbb{Q}$ as a topological vector space.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
Thus, every time we make a degree-$n$ extension of a field $F$, we're either just adding an extra limit that didn't exist before or we're moving from the original topology on $F$ to the finite product topology on $F^n$. And that determines our topology for $\widehat{\mathbb{Q}}$!1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers. So we can determine the topology of $\mathbb{C}$ by taking continuum many transcendental extensions of $\widehat{\mathbb{Q}}$ (algebraically completing them each time).
So now the question is to determine the topology of a transcendental extension $F[\alpha]$ of a topological field $F$. The set $\mathcal{F}$ of opens of $F$ whose closure includes $\alpha$ is clearly either a proper filter or the empty set3. If $\mathcal{F}$ is a proper filter,
Below are three examples/counterexamples of how this works:
2. We can set $\alpha$ to a previously undefined limit of a sequence; e.g. we could define $\alpha = 2.71828…$ if $2.71828…$ was not already present in the topological field $F$.
3. But note that we cannot define an infinitesimal, e.g. $\epsilon=0.0000…$, without making it topologically indistinguishable from 0 (any open set containing 0 would then contain $\epsilon$ as well), and thus making the whole topology indiscrete (by continuity of division).
Remaining questions
---
~~For this to become a satisfactory answer, there are some things that I'll have to figure out:~~
1. Can one prove that all metrics on $\mathbb{Q}$ are based on norms? (Of course, the Kuratowski embedding would homeomorphically embed $\mathbb{Q}$ into a normed vector space---but how do we get a *linear* embedding?)
2. Are there any norms on $\mathbb{Q}$ besides the Archimedean and $p$-adic norms?
~~3. What limitations are there on transcendental extensions over infinite-dimensional topologies (e.g. $\widehat{\mathbb{Q}}$ under the $p$-adic topology)?~~
---
1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer. Also, w.r.t. extending the topology on $F$ to the product topology on $F^n$, note that the finest possible topology on $F[\alpha]$, $\alpha$ algebraic of degree $n$, is symmetric under the Galois group of $\alpha$. Since it is symmetric, the Galois group actions are continuous automorphisms of $F[\alpha]$. Since the coefficients $a_i$ of any element $\sum_{i=0}^{n-1} a_i \alpha^i$ of $F[\alpha]$ can be extracted by summing Galois group actions, it follows that the $n$ projection maps from $F[\alpha]$ to $F$ are continuous, and hence that this topology on $F[\alpha]$ can be no finer than the product topology on $F^n$. Conversely, the product topology on $F^n$ clearly works as a topology for $F[\alpha]$, hence it is the finest possible topology on $F[\alpha]$. The only question remaining is whether there is a coarser topology possible on $F[\alpha]$ while still retaining the existence of a disjoint open containing
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
31. If $\mathcal{F}$ is the empty set (i.e. no open set in $F$ has a closure that includes $\alpha$, we get the finest possible topology on our extension. This is because $\alpha$ has an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).

Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
Since $\mathbb{Q}$ is second-countable and $T_3$, it is metrizable by Urysohn's theorem, for a proof of which see the nice exposition at the [nLab](https://ncatlab.org/nlab/show/Urysohn+metrization+theorem) ([archived copy](http://web.archive.org/web/2020/https://ncatlab.org/nlab/show/Urysohn+metrization+theorem)).
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a topological field has to follow either the usual topology or some $p$-adic topology, unfortunately requires the metric to be multiplicative, and I don't see how to show that given the kinds of metrics we get from metrization theorems. I don't even see how to show that we can necessarily get a norm on $\mathbb{Q}$ as a topological vector space.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
Thus, every time we make a degree-$n$ extension of a field $F$, we're either just adding an extra limit that didn't exist before or we're moving from the original topology on $F$ to the finite product topology on $F^n$. And that determines our topology for $\widehat{\mathbb{Q}}$!1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers. So we can determine the topology of $\mathbb{C}$ by taking continuum many transcendental extensions of $\widehat{\mathbb{Q}}$ (algebraically completing them each time).
So now the question is to determine the topology of a transcendental extension $F[\alpha]$ of a topological field $F$. The set $\mathcal{F}$ of opens of $F$ whose closure includes $\alpha$ is either a proper filter or the empty set3. Here are three examples of how this works.
1. If $\mathcal{F}$ is the empty set, then there are disjoint open sets that include $\alpha$ and any element in $F$. I believe one can also prove that $F$ is closed in $F[\alpha]$ and so there is a set including $\alpha$ that is disjoint from the whole of $F$.
2. We can set $\alpha$ to a previously undefined limit of a sequence; e.g. we could define $\alpha = 2.71828…$ if $2.71828…$ was not already present in the topological field $F$.
3. But note that we cannot define an infinitesimal, e.g. $\epsilon=0.0000…$, without making it topologically indistinguishable from 0 (any open set containing 0 would then contain $\epsilon$ as well), and thus making the whole topology indiscrete (by continuity of division).
Remaining questions
---
For this to become a satisfactory answer, there are some further things that I'll have to figure out:
1. Can one prove that all metrics on $\mathbb{Q}$ are based on norms? (Of course, the Kuratowski embedding would homeomorphically embed $\mathbb{Q}$ into a normed vector space---but how do we get a *linear* embedding?)
2. Are there any norms on $\mathbb{Q}$ besides the Archimedean and $p$-adic norms?
3. Is it possible to show that $F$ is closed in any field extension $F[\alpha]$?
4. What limitations are there on transcendental extensions over infinite-dimensional topologies (e.g. $\widehat{\mathbb{Q}}$ under the $p$-adic topology)?
---
1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer. Also, w.r.t. extending the topology on $F$ to the product topology on $F^n$, note that the finest possible topology on $F[\alpha]$, $\alpha$ algebraic of degree $n$, is symmetric under the Galois group of $\alpha$. Since it is symmetric, the Galois group actions are continuous automorphisms of $F[\alpha]$. Since the coefficients $a_i$ of any element $\sum_{i=0}^{n-1} a_i \alpha^i$ of $F[\alpha]$ can be extracted by summing Galois group actions, it follows that the $n$ projection maps from $F[\alpha]$ to $F$ are continuous, and hence that this topology on $F[\alpha]$ can be no finer than the product topology on $F^n$. Conversely, the product topology on $F^n$ clearly works as a topology for $F[\alpha]$, hence it is the finest possible topology on $F[\alpha]$. The only question remaining is whether there is a coarser topology possible on $F[\alpha]$ while still retaining the existence of a disjoint open containing
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
31. If $\mathcal{F}$ is the empty set (i.e. no open set in $F$ has a closure that includes $\alpha$, we get the finest possible topology on our extension. This is because $\alpha$ has an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).

#5: Post edited by $user avatar$ clemens‭ · 2026-03-08T21:03:02Z (6 months ago)

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Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
I believe that one can show that $\mathbb{Q}$ satisfies the stronger $T_{3½}$ separation axiom and hence is metrizable, but I do not at the moment see how.
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
And this basically determines our topology for $\widehat{\mathbb{Q}}$.1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers.
We have very wide latitude to define these transcendental numbers topologically. For example:
1. We can clearly define them as the topological limits of any net we want; e.g., we could redefine $\pi = 2.71828…, e=3.1415…$ (assuming $\pi$ and $e$ are transcendentally independent).3 \
2. We can also give these transcendental extensions a very fine topology, e.g. $\pi$ could have an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).
3. Transcendental numbers can also be chosen as infinitesimals, e.g. as the limit of the net of open sets $(0,\frac1n)$. They can even be chosen as arbitrary numbers within the ultrapower of the base field!
For transcendental extensions over a finite-dimensional topology (e.g. $\mathbb{R}$ with the usual topology, extended by adding a new transcendental number), we can use the $T_3$ property of our topological vector space to obtain a fundamental system of neighborhoods of the new transcendental number within the original space. E.g. call our transcendental number $\alpha$: then either $\alpha$ is within the closure of $[0,∞]$, or it has an open neighborhood disjoint from $[0,∞]$. In either case we can keep bisecting in this way to get our fundamental system, and every open set within $\mathbb{R}$, no matter how small, has to include the intersection with $\mathbb{R}$ of some element of this fundamental system.
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
1. Are there any topological field structures on $\mathbb{Q}$ besides the Archimedean topology and the $p$-adic topology?
2. What restrictions are there on the topology of transcendental extensions *in themselves* (so to speak) and not just in relation to the base field?
~~3. What can we say, topologically, about transcendental extensions over infinite-dimensional topologies, e.g. over $\widehat{\mathbb{Q}}$ under a $p$-adic topology?~~
---
~~1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer.~~
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
3 Obviously, $\pi$ and $e$ would lose their *topological* properties (e.g. that $e$ is the limit of $1$, $1+\frac12$, $1+\frac12+\frac16$, …) under this redefinition; but not, however, their *algebraic* properties, which are the ones under consideration here.

Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
Since $\mathbb{Q}$ is second-countable and $T_3$, it is metrizable by Urysohn's theorem, for a proof of which see the nice exposition at the [nLab](https://ncatlab.org/nlab/show/Urysohn+metrization+theorem) ([archived copy](http://web.archive.org/web/2020/https://ncatlab.org/nlab/show/Urysohn+metrization+theorem)).
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a topological field has to follow either the usual topology or some $p$-adic topology, unfortunately requires the metric to be multiplicative, and I don't see how to show that given the kinds of metrics we get from metrization theorems. I don't even see how to show that we can necessarily get a norm on $\mathbb{Q}$ as a topological vector space.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
Thus, every time we make a degree-$n$ extension of a field $F$, we're either just adding an extra limit that didn't exist before or we're moving from the original topology on $F$ to the finite product topology on $F^n$. And that determines our topology for $\widehat{\mathbb{Q}}$!1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers. So we can determine the topology of $\mathbb{C}$ by taking continuum many transcendental extensions of $\widehat{\mathbb{Q}}$ (algebraically completing them each time).
So now the question is to determine the topology of a transcendental extension $F[\alpha]$ of a topological field $F$. The set $\mathcal{F}$ of opens of $F$ whose closure includes $\alpha$ is clearly either a proper filter or the empty set3. If $\mathcal{F}$ is a proper filter,
Below are three examples/counterexamples of how this works:
2. We can set $\alpha$ to a previously undefined limit of a sequence; e.g. we could define $\alpha = 2.71828…$ if $2.71828…$ was not already present in the topological field $F$.
3. But note that we cannot define an infinitesimal, e.g. $\epsilon=0.0000…$, without making it topologically indistinguishable from 0 (any open set containing 0 would then contain $\epsilon$ as well), and thus making the whole topology indiscrete (by continuity of division).
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
1. Can one prove that all metrics on $\mathbb{Q}$ are based on norms? (Of course, the Kuratowski embedding would homeomorphically embed $\mathbb{Q}$ into a normed vector space---but how do we get a *linear* embedding?)
2. Are there any norms on $\mathbb{Q}$ besides the Archimedean and $p$-adic norms?
3. What limitations are there on transcendental extensions over infinite-dimensional topologies (e.g. $\widehat{\mathbb{Q}}$ under the $p$-adic topology)?
---
1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer. Also, w.r.t. extending the topology on $F$ to the product topology on $F^n$, note that the finest possible topology on $F[\alpha]$, $\alpha$ algebraic of degree $n$, is symmetric under the Galois group of $\alpha$. Since it is symmetric, the Galois group actions are continuous automorphisms of $F[\alpha]$. Since the coefficients $a_i$ of any element $\sum_{i=0}^{n-1} a_i \alpha^i$ of $F[\alpha]$ can be extracted by summing Galois group actions, it follows that the $n$ projection maps from $F[\alpha]$ to $F$ are continuous, and hence that this topology on $F[\alpha]$ can be no finer than the product topology on $F^n$. Conversely, the product topology on $F^n$ clearly works as a topology for $F[\alpha]$, hence it is the finest possible topology on $F[\alpha]$. The only question remaining is whether there is a coarser topology possible on $F[\alpha]$ while still retaining the existence of a disjoint open containing
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
31. If $\mathcal{F}$ is the empty set (i.e. no open set in $F$ has a closure that includes $\alpha$, we get the finest possible topology on our extension. This is because $\alpha$ has an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).

#4: Post edited by $user avatar$ clemens‭ · 2026-03-02T19:40:31Z (7 months ago)

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Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
I believe that one can show that $\mathbb{Q}$ satisfies the stronger $T_{3½}$ separation axiom and hence is metrizable, but I do not at the moment see how.
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
And this basically determines our topology for $\widehat{\mathbb{Q}}$.1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers.
We have very wide latitude to define these transcendental numbers topologically. For example:
1. We can clearly define them as the topological limits of any net we want; e.g., we could redefine $\pi = 2.71828…, e=3.1415…$ (assuming $\pi$ and $e$ are transcendentally independent).3 \
2. We can also give these transcendental extensions a very fine topology, e.g. $\pi$ could have an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).
3. Transcendental numbers can also be chosen as infinitesimals, e.g. as the limit of the net of open sets $(0,\frac1n)$. They can even be chosen as arbitrary numbers within the ultrapower of the base field!
For transcendental extensions over a finite-dimensional topology (e.g. $\mathbb{R}$ with the usual topology, extended by adding a new transcendental number), we can use the $T_3$ property of our topological vector space to obtain a fundamental system of neighborhoods of the new transcendental number within the original space. E.g. call our transcendental number $\alpha$: then either $\alpha$ is within the closure of $[0,∞]$, or it has an open neighborhood disjoint from $[0,∞]$. In either case we can keep bisecting in this way to get our fundamental system, and every open set within $\mathbb{R}$, no matter how small, has to include the intersection with $\mathbb{R}$ of some element of this fundamental system.
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
~~1. Are there any topological fields isomorphic to $\mathbb{Q}$ that are different from the Archimedean topology and the $p$-adic topology?~~
2. What restrictions are there on the topology of transcendental extensions *in themselves* (so to speak) and not just in relation to the base field?
3. What can we say, topologically, about transcendental extensions over infinite-dimensional topologies, e.g. over $\widehat{\mathbb{Q}}$ under a $p$-adic topology?
---
1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer.
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
3 Obviously, $\pi$ and $e$ would lose their *topological* properties (e.g. that $e$ is the limit of $1$, $1+\frac12$, $1+\frac12+\frac16$, …) under this redefinition; but not, however, their *algebraic* properties, which are the ones under consideration here.

Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
I believe that one can show that $\mathbb{Q}$ satisfies the stronger $T_{3½}$ separation axiom and hence is metrizable, but I do not at the moment see how.
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
And this basically determines our topology for $\widehat{\mathbb{Q}}$.1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers.
We have very wide latitude to define these transcendental numbers topologically. For example:
1. We can clearly define them as the topological limits of any net we want; e.g., we could redefine $\pi = 2.71828…, e=3.1415…$ (assuming $\pi$ and $e$ are transcendentally independent).3 \
2. We can also give these transcendental extensions a very fine topology, e.g. $\pi$ could have an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).
3. Transcendental numbers can also be chosen as infinitesimals, e.g. as the limit of the net of open sets $(0,\frac1n)$. They can even be chosen as arbitrary numbers within the ultrapower of the base field!
For transcendental extensions over a finite-dimensional topology (e.g. $\mathbb{R}$ with the usual topology, extended by adding a new transcendental number), we can use the $T_3$ property of our topological vector space to obtain a fundamental system of neighborhoods of the new transcendental number within the original space. E.g. call our transcendental number $\alpha$: then either $\alpha$ is within the closure of $[0,∞]$, or it has an open neighborhood disjoint from $[0,∞]$. In either case we can keep bisecting in this way to get our fundamental system, and every open set within $\mathbb{R}$, no matter how small, has to include the intersection with $\mathbb{R}$ of some element of this fundamental system.
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
1. Are there any topological field structures on $\mathbb{Q}$ besides the Archimedean topology and the $p$-adic topology?
2. What restrictions are there on the topology of transcendental extensions *in themselves* (so to speak) and not just in relation to the base field?
3. What can we say, topologically, about transcendental extensions over infinite-dimensional topologies, e.g. over $\widehat{\mathbb{Q}}$ under a $p$-adic topology?
---
1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer.
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
3 Obviously, $\pi$ and $e$ would lose their *topological* properties (e.g. that $e$ is the limit of $1$, $1+\frac12$, $1+\frac12+\frac16$, …) under this redefinition; but not, however, their *algebraic* properties, which are the ones under consideration here.

#3: Post edited by $user avatar$ clemens‭ · 2026-03-02T19:27:59Z (7 months ago)
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Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
~~I believe that one can show that $\mathbb{Q}$ satisfied the stronger $T_{3½}$ separation axiom and hence is metrizable, but I do not at the moment see how.~~
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
~~And this basically determines our topology for $\overline{\mathbb{Q}}$.~~
Transcendental extensions
---
~~Next, the OP asks us to define a topology not only on $\overline{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$1.~~
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\overline{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers.
We have very wide latitude to define these transcendental numbers topologically. For example:
1. We can clearly define them as the topological limits of any net we want; e.g., we could redefine $\pi = 2.71828…, e=3.1415…$ (assuming $\pi$ and $e$ are transcendentally independent).2 \
2. We can also give these transcendental extensions a very fine topology, e.g. $\pi$ could have an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).
3. Transcendental numbers can also be chosen as infinitesimals, e.g. as the limit of the net of open sets $(0,\frac1n)$. They can even be chosen as arbitrary numbers within the ultrapower of the base field!
For transcendental extensions over a finite-dimensional topology (e.g. $\mathbb{R}$ with the usual topology, extended by adding a new transcendental number), we can use the $T_3$ property of our topological vector space to obtain a fundamental system of neighborhoods of the new transcendental number within the original space. E.g. call our transcendental number $\alpha$: then either $\alpha$ is within the closure of $[0,∞]$, or it has an open neighborhood disjoint from $[0,∞]$. In either case we can keep bisecting in this way to get our fundamental system, and every open set within $\mathbb{R}$, no matter how small, has to include the intersection with $\mathbb{R}$ of some element of this fundamental system.
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
1. Are there any topological fields isomorphic to $\mathbb{Q}$ that are different from the Archimedean topology and the $p$-adic topology?
2. What restrictions are there on the topology of transcendental extensions *in themselves* (so to speak) and not just in relation to the base field?
~~3. What can we say, topologically, about transcendental extensions over infinite-dimensional topologies, e.g. over $\overline{\mathbb{Q}}$ under a $p$-adic topology?~~
---
~~1Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.~~
2 Obviously, $\pi$ and $e$ would lose their *topological* properties (e.g. that $e$ is the limit of $1$, $1+\frac12$, $1+\frac12+\frac16$, …) under this redefinition; but not, however, their *algebraic* properties, which are the ones under consideration here.

Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
I believe that one can show that $\mathbb{Q}$ satisfies the stronger $T_{3½}$ separation axiom and hence is metrizable, but I do not at the moment see how.
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
And this basically determines our topology for $\widehat{\mathbb{Q}}$.1
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\widehat{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$2.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\widehat{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers.
We have very wide latitude to define these transcendental numbers topologically. For example:
1. We can clearly define them as the topological limits of any net we want; e.g., we could redefine $\pi = 2.71828…, e=3.1415…$ (assuming $\pi$ and $e$ are transcendentally independent).3 \
2. We can also give these transcendental extensions a very fine topology, e.g. $\pi$ could have an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).
3. Transcendental numbers can also be chosen as infinitesimals, e.g. as the limit of the net of open sets $(0,\frac1n)$. They can even be chosen as arbitrary numbers within the ultrapower of the base field!
For transcendental extensions over a finite-dimensional topology (e.g. $\mathbb{R}$ with the usual topology, extended by adding a new transcendental number), we can use the $T_3$ property of our topological vector space to obtain a fundamental system of neighborhoods of the new transcendental number within the original space. E.g. call our transcendental number $\alpha$: then either $\alpha$ is within the closure of $[0,∞]$, or it has an open neighborhood disjoint from $[0,∞]$. In either case we can keep bisecting in this way to get our fundamental system, and every open set within $\mathbb{R}$, no matter how small, has to include the intersection with $\mathbb{R}$ of some element of this fundamental system.
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
1. Are there any topological fields isomorphic to $\mathbb{Q}$ that are different from the Archimedean topology and the $p$-adic topology?
2. What restrictions are there on the topology of transcendental extensions *in themselves* (so to speak) and not just in relation to the base field?
3. What can we say, topologically, about transcendental extensions over infinite-dimensional topologies, e.g. over $\widehat{\mathbb{Q}}$ under a $p$-adic topology?
---
1 I have to use the \widehat $\LaTeX$ command instead of the \overline command because of a bug in the MathJax renderer.
2Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
3 Obviously, $\pi$ and $e$ would lose their *topological* properties (e.g. that $e$ is the limit of $1$, $1+\frac12$, $1+\frac12+\frac16$, …) under this redefinition; but not, however, their *algebraic* properties, which are the ones under consideration here.

#2: Post edited by $user avatar$ clemens‭ · 2026-03-02T05:58:21Z (7 months ago)

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~~Here's a tentative answer that I hope will be useful. Any corrections are of course very welcome.~~
A topology on $\mathbb{Q}$
---
~~Any topological vector space has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. .~~
Hence $\mathbb{Q}$ is metrizable by Urysohn's theorem.
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
And this basically determines our topology for $\overline{\mathbb{Q}}$.
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\overline{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$1.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\overline{\mathbb{Q}}$ is that we've added an uncountable number of transcendentally independent numbers and taken the algebraic closure of the result.
---
~~1Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.~~

Here's an incomplete answer that I hope will be useful. The basic question of *whether* there are any non-standard ways to give a topological field structure to $\mathbb{R}$ is of course answerable in the positive, but I tried to answer the deeper question of characterizing the various ways in which $\mathbb{R}$ can be given a topological field structure. Any corrections are of course very welcome.
A topology on $\mathbb{Q}$
---
Any topological abelian group $\mathcal{T}$ has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. Also, the set $\{(x,C + x): x \in \mathcal{T} \}$ is closed for any closed $C$, so $\mathcal{T}$ is also $T_3$.
I believe that one can show that $\mathbb{Q}$ satisfied the stronger $T_{3½}$ separation axiom and hence is metrizable, but I do not at the moment see how.
Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.
Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---
We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology.
Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction.
And this basically determines our topology for $\overline{\mathbb{Q}}$.
Transcendental extensions
---
Next, the OP asks us to define a topology not only on $\overline{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$1.
Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\overline{\mathbb{Q}}$ is that $\mathbb{C}$ includes an uncountable number of algebraically independent transcendental numbers.
We have very wide latitude to define these transcendental numbers topologically. For example:
1. We can clearly define them as the topological limits of any net we want; e.g., we could redefine $\pi = 2.71828…, e=3.1415…$ (assuming $\pi$ and $e$ are transcendentally independent).2 \
2. We can also give these transcendental extensions a very fine topology, e.g. $\pi$ could have an open neighborhood disjoint from all rational or algebraic numbers (much as $\sqrt2$, in the 3-adic topology on $\mathbb{Q}[\sqrt2]$, has an open neighborhood disjoint from all numbers that do not involve $\sqrt2$).
3. Transcendental numbers can also be chosen as infinitesimals, e.g. as the limit of the net of open sets $(0,\frac1n)$. They can even be chosen as arbitrary numbers within the ultrapower of the base field!
For transcendental extensions over a finite-dimensional topology (e.g. $\mathbb{R}$ with the usual topology, extended by adding a new transcendental number), we can use the $T_3$ property of our topological vector space to obtain a fundamental system of neighborhoods of the new transcendental number within the original space. E.g. call our transcendental number $\alpha$: then either $\alpha$ is within the closure of $[0,∞]$, or it has an open neighborhood disjoint from $[0,∞]$. In either case we can keep bisecting in this way to get our fundamental system, and every open set within $\mathbb{R}$, no matter how small, has to include the intersection with $\mathbb{R}$ of some element of this fundamental system.
Remaining questions
---
For this to become a satisfactory answer, there are some things that I'll have to figure out:
1. Are there any topological fields isomorphic to $\mathbb{Q}$ that are different from the Archimedean topology and the $p$-adic topology?
2. What restrictions are there on the topology of transcendental extensions *in themselves* (so to speak) and not just in relation to the base field?
3. What can we say, topologically, about transcendental extensions over infinite-dimensional topologies, e.g. over $\overline{\mathbb{Q}}$ under a $p$-adic topology?
---
1Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.
2 Obviously, $\pi$ and $e$ would lose their *topological* properties (e.g. that $e$ is the limit of $1$, $1+\frac12$, $1+\frac12+\frac16$, …) under this redefinition; but not, however, their *algebraic* properties, which are the ones under consideration here.

#1: Initial revision by $user avatar$ clemens‭ · 2026-03-02T05:17:48Z (7 months ago)

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Here's a tentative answer that I hope will be useful. Any corrections are of course very welcome.

A topology on $\mathbb{Q}$
---

Any topological vector space has closed diagonal (because the $-$ function is continuous) and hence is Hausdorff. . 

Hence $\mathbb{Q}$ is metrizable by Urysohn's theorem.

Ostrowski's theorem, which would show that $\mathbb{Q}$ as a normed vector space has to follow either the usual topology or some $p$-adic topology, unfortunately requires multiplicativity of the metric, and I don't see how to show that given the kinds of metrics we get from metrization theorems.

Extending $\mathbb{Q}$ to $\overline{\mathbb{Q}}$
---

We can extend $\mathbb{Q}$ by adding extra limits of nets therein. For instance, we can extend $\mathbb{Q}$ to $\mathbb{Q}[\sqrt2]$ by defining $\sqrt2 = 1.414213562…$ in the usual topology or $\sqrt2=…2 0 1 1 2 6 6 4 2 1 2 1 6 2 1 3$ in the 7-adic topology. 

Since polynomials are continuous, this comes with an important restriction: we cannot make the root of any polynomial into the limit of the "wrong" net. E.g. if we tried to take $\sqrt2 = 1.732…$, we would get that $2 = \sqrt2^2 = (1.732…)^2 = 3$, a contradiction. 

And this basically determines our topology for $\overline{\mathbb{Q}}$.

Transcendental extensions
---

Next, the OP asks us to define a topology not only on $\overline{\mathbb{Q}}$ which contains roots but also on $\mathbb{R}$, or equivalently, on $\mathbb{C}$<sup>1</sup>.

Well, algebraically speaking, the only difference between $\mathbb{C}$ and its subset $\overline{\mathbb{Q}}$ is that we've added an uncountable number of transcendentally independent numbers and taken the algebraic closure of the result.





---
<sup>1</sup>Since any topology on $\mathbb{R}$ induces a topology on $\mathbb{C}$, we might as well define our topology on $\mathbb{C}$.

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