Post History
#2: Post edited
by
celtschk
·
2025-11-14T04:59:58Z (10 months ago)
Added: My concrete idea from the question does not work
- The answer is yes. The key to this is found in the [Wikipedia article about p-adic numbers:](https://en.wikipedia.org/wiki/P-adic_number?utm_source=chatgpt.com#Algebraic_closure)
- > $\mathbb {C} _{p}$ 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 $\mathbb R$.
- The answer is yes. The key to this is found in the [Wikipedia article about p-adic numbers:](https://en.wikipedia.org/wiki/P-adic_number?utm_source=chatgpt.com#Algebraic_closure)
- > $\mathbb {C} _{p}$ 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).
#1: Initial revision
by
celtschk
·
2025-11-13T05:34:35Z (10 months ago)
The answer is yes. The key to this is found in the [Wikipedia article about p-adic numbers:](https://en.wikipedia.org/wiki/P-adic_number?utm_source=chatgpt.com#Algebraic_closure)
> $\mathbb {C} _{p}$ 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 $\mathbb R$.