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In wikipedia, normed vector space is defined as a vector space over a subfield of $\mathbb{C}$ equipped with a norm.

However, Banach space is defined as a complete normed space over $\mathbb{R}$ or $\mathbb{C}$.

Is there a reason that we consider only real or complex Banach space?

Since every normed space over a subfield $F$ of $\mathbb{C}$ can be completed to a complete normed space over $F$, I don't get why we are not considering these complete normed spaces over $F$.

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    $\begingroup$ I guess that it is not too hard to show that if $V$ is a complete $\Bbb{Q}$ vector space, one can extend the scalar multiplication uniquely continuously to $\Bbb{R}\times V\to V$, so that $V$ is also a $\Bbb{R}$ vector space. Hence, we assume this to begin with. Also, if the vector space is complete, it is natural to assume that the underlying field is complete too. $\endgroup$
    – PhoemueX
    Jan 2 '15 at 14:56
  • $\begingroup$ @PhoemueX thank you :) would you write that as an answer? $\endgroup$
    – Rubertos
    Jan 2 '15 at 14:58
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    $\begingroup$ In a different direction, we do not consider Banach spaces only over $\mathbf R$ and $\mathbf C$. Banach spaces over the $p$-adic numbers and other non-archimedean complete fields are important as well, at least in number theory. Google "non-archimedean functional analysis." $\endgroup$
    – KCd
    Jan 2 '15 at 15:01
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It is not too hard to show that if $V$ is a complete $\Bbb{Q}$ vector space, one can extend the scalar multiplication uniquely continuously to $\Bbb{R}\times V\to V$, so that $V$ is also a $\Bbb{R}$ vector space. Hence, we assume this to begin with.

Also, if the vector space is complete, it is natural to assume that the underlying field is complete too.

EDIT: Here, I consider $\Bbb{Q}$ to be equipped with the usual absolute value $|x|=\max \{x,-x\}$ and assume that the norm $\Vert \cdot \Vert$ on $V$ satisfies $\Vert \alpha x \Vert = |\alpha| \cdot \Vert x\Vert$ for $x\in V$ and $\alpha \in \Bbb{Q}$.

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  • $\begingroup$ Be more careful by specifying what absolute value you are putting on $\mathbf Q$ when you speak about a "normed $\mathbf Q$-vector space." Otherwise it can sound like you are suggesting there is some kind of natural/canonical multiplication $\mathbf R \times \mathbf Q_p \rightarrow \mathbf Q_p$ by taking for $V$ the field of $p$-adic numbers, which is a complete vector space over $\mathbf Q$. $\endgroup$
    – KCd
    Jan 2 '15 at 15:17
  • $\begingroup$ @KCd: I edited the post. Are you satisfied with my formulation? $\endgroup$
    – PhoemueX
    Jan 2 '15 at 15:38
  • $\begingroup$ Yes, it looks good. $\endgroup$
    – KCd
    Jan 2 '15 at 16:17

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