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.
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