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This is an exercise problem from a textbook on Functional analysis.

Show that the discrete metric on a vector space $X\ne \{0\}$ cannot be obtained from a norm.

If the field over which $X$ is a vector space is $\mathbb R$ or $\mathbb C$, then for any $t\in X-\{0\}$, we have $d(t,0)=\|t\|=1$.

Note that $d(2t,0)=\|2t\|=2\|t\|=2$, which is a contradiction.

But how to prove the statement in case of arbitrary fields? I have also dropped comment here to an older post on the same exercise. But the post does not seem to be considering arbitrary fields.

Koro
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    Which textbook is this? Most textbooks of functional analysis do not consider fields other than the real and complex numbers – J. De Ro Dec 21 '22 at 11:17
  • @QuantumSpace: It's Kreyzig's book on functional analysis. – Koro Dec 21 '22 at 11:20
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    @Koro How do you define a norm on an $F$-vector space $X$, where $F$ is a field? For example, you should have something like $|\alpha x |= |\alpha||x|$ which suggests that your field should at least have an "absolute value function" $F \to [0,\infty[$? – J. De Ro Dec 21 '22 at 11:21
  • @QuantumSpace: I think the norm is defined only on real or complex vector spaces (i.e. vector space over R or over C). Is my understanding correct? Do we define norm over arbitrary fields? – Koro Dec 21 '22 at 11:23
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    That's precisely the point of my comments. We usually don't, and if we do we need more structure on the field. Look up "Archimedian field". – J. De Ro Dec 21 '22 at 11:29
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    @QuantumSpace Valued fields are a thing and so are normed vector spaces over valued fields. You don't need an Archimedian field to do functional analysis. I agree that it is unlikely that a standard functional analysis text considers this generality. But saying that you need an Archimedian field is not correct. By the way, it's possible for a general valued field that the discrete metric is induced by a norm, but that's only possible if the absolute value is trivial. – Lukas Heger Dec 21 '22 at 12:02
  • @LukasHeger You are probably right. I never use fields other than $\mathbb{C}$. My point was just that you need extra structure on the field. – J. De Ro Dec 21 '22 at 16:05
  • @QuantumSpace: Thanks a lot. I understand :). – Koro Dec 22 '22 at 05:28
  • @quantumspace I absolutely agree that you need extra structure, I was just pointing out what kind of extra structure you need. Your first guess that one needs some kind of "absolute value function" was spot-on, in fact. – Lukas Heger Dec 22 '22 at 23:43

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