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Topic: What is the intuitive meaning of "non-Archimedean" for a valued field?
Replies: 11   Last Post: May 15, 2013 4:29 PM

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William Elliot

Posts: 2,637
Registered: 1/8/12
Re: What is the intuitive meaning of "non-Archimedean" for a valued

Posted: May 13, 2013 10:27 PM
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On Mon, 13 May 2013, mike3 wrote:

> I'm curious about this. The "Archimedean property" for an _ordered_
> field F means that given any positive elements a and b in F, with a <
> b, then there exists a natural number n such that na < b.

That is trival. If a < b, then 1a < b and the Archimedean property
always holds.

> Intuitively, this means F has no "infinitely big" or "infinitely small"
> elements.

> We could also say that "given any positive element a, then
> there is a natural number n such that na > 1". If the property fails,
> then F contains "infinitely small" elements.

What does infinitely large and infinitely small mean?

> Now, there is an analogous property for non-ordered, "valued" fields
> (fields with an "absolute value" function added). The "Archimedean
> property" here means that given any nonzero element a e F, that there
> exists a natural number n such that |na| > 1. But what, intuitively,
> does it mean when this property fails?

That there's some element a, for which the sequence
(|na|)_n is bounded by 1.

> In that case, there aren't any elements with "infinitely small but
> non-zero" absolute value since the absolute value functions are usually
> taken as real-valued, and the reals are Archimedean (as an ordered
> field). Instead, what happens for such a real-valued absolute value is
> that the triangle inequality strengthens to |a + b| <= max(|a|, |b|) and
> not just |a + b| <= |a| + | b|. This causes the space to behave really
> weirdly(*). But what is the _intuition_ here, and how does this notion
> relate, if at all, to the first one? Especially considering I see in
> papers like this:

It seems such a norm would give rise to an ultra metric.

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