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The $p$-adic norm of an integer $k$ is $||k||_p := p^{-\alpha}$ where $p^\alpha$ is the largest power of $p$ that divides $k$.

This function can be easily extended to the rationals as $||a/b||_p := ||a||_p/||b||_p$ (notice that's well defined).

Apparently there is this Chevalley's Theorem which states that this norm can be even extended to the real numbers preserving several nice proprieties.

Is this extended norm continuous?

Alma Arjuna
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    I'm not sure what theorem of Chevalley you're referring to. Do you have a reference or something pointing to where you heard this? I don't see any natural way to extend the $p$-adic norm to the real numbers. – Daniel Hast Oct 06 '22 at 00:38
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    Please clarify what exact map's continuity you are interested in, specifying both domain and codomain as topological spaces. – Torsten Schoeneberg Oct 06 '22 at 01:04
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    Here's a pdf on Monsky's thorem which contains a proof of a "Chevalley's theorem" which extends the p-adic absolute value to $\mathbb{R}$. I'd be willing to bet this is what's being asked about. I suppose there is a bit of an ambiguity in that for the sake of Monsky's theorem you only ever need a finite extension and don't need to extend it to all of $\mathbb{R}$. (I haven't thought about whether that changes the answer or not) https://math.osu.edu/sites/math.osu.edu/files/ferre_MonskyThm_2016.pdf – Merosity Oct 06 '22 at 01:22
  • "that this norm can be even extended to the real numbers preserving several nice [properties]". I am not aware of any "nice properties" of those (ridiculously infinitely many) non-explicit extensions beyond satisfying the definition of absolute values. Their existence relies on the axiom of choice. Compare https://math.stackexchange.com/q/2563024/96384, https://math.stackexchange.com/q/338148/96384, https://math.stackexchange.com/a/4007515/96384, https://math.stackexchange.com/q/1348581/96384 (note comments under answers). – Torsten Schoeneberg Oct 06 '22 at 17:49

1 Answers1

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It is not.

If it were continuous, it would be continuous at $0$. Then, there would exist $\delta>0$ such that if $|x|<\delta$ then $|||x||_p- ||0||_p|=||x||_p<1$.

However, this is impossible since $||\frac{1}{p^k}||_p = \frac{||1||_p}{||p^k||_p}=\frac{p^{-0}}{p^{-k}}=p^k \to \infty$ and for $k$ big enough we will have that $|\frac{1}{p^k}|<\delta$.

Nicolas Agote
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    This proves that the $p$-adic norm $||_p\colon\mathbb Q\to\mathbb R$ is discontinuous, where $\mathbb Q$ is given the subspace topology induced from $\mathbb R$. In fact, this function is nowhere continuous! – Kenta S Oct 06 '22 at 00:56