Are there other topologies on $\mathbb R$ that make it a topological field?
As is well known, $\mathbb R$ with the standard topology is a topological field. It is also not hard to check that the discrete and the indiscrete topology on $\mathbb R$ result in a topological field, simply from the fact that all functions from a discrete topology are continuous, as are all functions to an indiscrete topology.
However I wonder if there are other topologies that make $\mathbb R$ into a topological field, in particular other topologies that can be easily written down.
Here's one idea: Let's define a set as open if its intersection with the rational numbers agrees with the intersection of an open set of the standard topology with the rational numbers. This is easily seen to be a topology, however I'm not sure if it makes all operations continuous.
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The answer is yes. The key to this is found in the Wikipedia article about p-adic numbers:
$\mathbb {C} _{p}$ and $\mathbb {C}$ are isomorphic as rings, so we may regard $\mathbb {C} _{p}$ as $\mathbb {C}$ endowed with an exotic metric. The proof of existence of such a field isomorphism relies on the axiom of choice, and does not provide an explicit example of such an isomorphism (that is, it is not constructive).
Here $\mathbb C_p$ is the metric completion of the algebraic closure of the field of $p$-adic numbers.
Obviously that exotic metric can be restricted to $\mathbb R$ and therefore gives rise to a non-standard topology on $\mathbb R$.
However my idea in the question does not lead to a topological field because it is not translation invariant. An easy way to see that is to note that in a topological field either all singletons are open, or no singletons are open, since translations can map any singleton to any other. However in the topology from my idea, rational singletons are not open (there's no open set in the standard topology that contains exactly one rational number), but irrational singletons are (the intersection of an irrational singleton with the rational numbers obviously is empty, and therefore open).
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This is only half of an answer but here we go:
First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies. And to separate algebra from topology.
Name the topology as $\mathcal T$.
Addition and multiplication can be seen as translation and scaling.
For any field $X$, let $a\in X$ and $a\neq 0$.
Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all these functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
Trouble here: I missed that from my assumptions above, it does not follow (or at least, I can't see it and missed to show it) that the multiplicative inverse function $a \mapsto a^{-1}$ also is continuous.
This is co-didact, which I read as "teaching each other", so what's wrong and why didn't anyone notice and tell? When I misspelled "Turing machine" as "Touring machine" the coment was promptly there.
So, if you throw in any collection of sets $\mathcal A\subset \mathcal P(X)$ you want to call open, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
This is where my wisdom comes to an end and why I call it half of an answer.
I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
To explain a bit what is in my mind when I say "collapse":
Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, every set with a single element will be open, and you end up with the discrete topology.
Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
Having any non-empty final set $A\in\mathcal A$ (assuming $A$ is not translation invariant, as can happen for final fields $X$; missed that case at first and nobody noticed! ) in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
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 (archived copy).
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.
- 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 $F$.
- 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 $F$.
- 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:
- 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?)
- Are there any norms on $\mathbb{Q}$ besides the Archimedean and $p$-adic norms?
- Is it possible to show that $F$ is closed in any field extension $F[\alpha]$?
- 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 $\sqrt2$).
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