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My question: can we define Banach space over the field $\mathbb Q$ instead of over the usual $\mathbb R$ and $\mathbb C$? Will $\mathbb R$ be a Banach space over $\mathbb Q$ if it can be done?

The reason why I am asking this is because I have read that every separable Banach space does not have Schauder basis.

Sampah
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Rajat
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  • What do you mean by "replace the field choice by Q replacing R/C"? – 5xum Mar 28 '22 at 09:08
  • So whenever I see the defination of Banach space it's always on a field, and it's mostly either R or C(Real or Complex) if I want the field to be Q(rational) will i still be able to define Banach space? – Rajat Mar 28 '22 at 09:11
  • And if I do that will R be a banach space over the field Q or not? – Rajat Mar 28 '22 at 09:13

2 Answers2

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The problem with doing analysis over $\mathbb{Q}$, is that $\mathbb{Q}$ lacks the completeness property: not every Cauchy sequence in $\mathbb{Q}$ converges.

This makes it pretty much impossible to do analysis over $\mathbb{Q}$, after all "natural" theorems become false: For instance it’s possible to have a monotonic bounded sequence which does not converge, or have a bounded set with no LUB, or have a continuous function which fails IVT, etc.

Vivaan Daga
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  • For the sake of completion (especially meant to the OP of the question), there is no non-trivial norm on $\mathbb Q$ such that $\mathbb Q$ is a complete normed vector space over itself. In case he/she or others wonders we should introduce a different norm on the rationals. – Sampah Apr 01 '22 at 12:06
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Banach spaces, by definition, are normed vector spaces.

And by definition, normed vector spaces are spaces over $\mathbb C$ or $\mathbb R$ (see here). Therefore, you cannot define a Banach space over $\mathbb Q$.

5xum
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    The natural follow up question is "What goes wrong with a normed space over $\mathbb{Q}$?" – Korf Mar 28 '22 at 09:23
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    @Korf $\mathbb{Q}$ lacks completeness. eg in $\mathbb{Q}^2$ with an "orthonormal" basis $(e_1, e_2)$ the vector with coordinates $(1, 1)$ is expected to have norm $\sqrt{2}$ but $\sqrt{2} \notin \mathbb{Q}$... – Olivier Roche Mar 28 '22 at 09:37