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If $X \neq \{ 0\}$ is a vector space. How does one go about showing that the discrete metric on $X$ cannot be obtained from any norm on $X$?

I know this is because $0$ does not lie in $X$, but I am having problems. Formalizing a proof for this.

This is also my final question for some time, after this I will reread the answers, and not stop until I can finally understand these strange spaces.

M Turgeon
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N3buchadnezzar
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4 Answers4

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HINT: Suppose that the norm $\|\cdot\|$ generates the discrete topology on $X$. Then there is an $\epsilon>0$ such that $\{x\in X:\|x\|<\epsilon\}=\{0\}$. By hypothesis $X$ contains at least one non-zero vector $y$. Let $\alpha=\|y\|>0$. Where is the vector $\dfrac{\epsilon}{2\alpha}y$?

Brian M. Scott
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  • Hmm. You are using discrete topology where the question asker said discrete metric, but I won't downvote because I think this is interesting itself. – Ben Millwood Sep 21 '12 at 01:29
  • @Ben: I’m a bit bemused that anyone would even consider downvoting on that basis: The discrete metric generates the discrete topology, so showing that the latter is not generated by a norm is a perfectly reasonable way to show that the former isn’t (and more besides). – Brian M. Scott Sep 21 '12 at 06:42
  • Oh, I hadn't thought of that. I had assumed that you had answered a genuinely different question. (Still, the same question for the discrete metric is more obvious in my view, cf. the other answer) – Ben Millwood Sep 21 '12 at 10:55
  • @Ben: What’s obvious depends so much on one’s background. A functional analyst would probably agree with you; I’m a set-theoretic topologist, so it’s not really surprising that I find the topological solution much more obvious. – Brian M. Scott Sep 21 '12 at 10:58
  • @BrianM.Scott:$\dfrac{\epsilon}{2\alpha}y$ will be in ${x\in X:|x|<\epsilon}={0}$,Which is an absurd equation... – Styles Jul 06 '19 at 08:36
  • I have one small question: Is the statement in OP true in case of X over any field? If the field is R or C, then clearly for any non zero x, we have $d(x,0)=1$ and ||2x||=2||x||=2= d(2x,0)=1, a contradiction. But if the field is not R or C, then how can one say that $\frac {\epsilon}{2\alpha}$ is a scalar? – Koro Dec 21 '22 at 11:03
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    @Koro: the vectors spaces studied in functional analysis are over the real or the complex fields. – Mittens Dec 21 '22 at 15:29
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    @Koro: To define norms and in general the concept of topological vector space over arbitrary field you need to define a valuation (absolute value or modulus) on a field. The definition of norm is similar to the one in complex (real) normed spaces. Things may not very be well behaved however, specially for finite fields. If you are interest in that, here is areference: C.Perez-Garcia, W.H.Schikhof, Locally Convex Spaces over non-Arquimedean Valued Fields, Cambridge Studies in Advanced Mathematics (2010). – Mittens Dec 21 '22 at 16:06
  • @OliverDíaz: Thanks for the help; I would not have been able to offer a reference. – Brian M. Scott Dec 21 '22 at 19:47
  • @OliverDíaz: Thanks a lot :-). – Koro Dec 21 '22 at 20:19
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You know that the discrete metric only takes values of $1$ and $0$. Now suppose it comes from some norm $||.||$. Then for any $\alpha$ in the underlying field of your vector space and $x,y \in X$, you must have that

$$\lVert\alpha(x-y)\rVert = \lvert\alpha\rvert\,\lVert x-y\rVert.$$

But now $||x-y||$ is a fixed number and I can make $\alpha$ arbitrarily large and consequently the discrete metric does not come from any norm on $X$.

Ben Millwood
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  • I guessed you meant $|\alpha|$ rather than $\alpha$. I also played around a bit with your $\LaTeX$ in the hope of making that a bit clearer. – Ben Millwood Sep 21 '12 at 13:50
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Suppose$(X,d)$ is a vector space with the discrete metric. If$X$ is a normed space, then for ∀ $x≠y ∈ X$,$α≠0 ∈K, αx≠αy$.So then $\lVert\alpha x-\alpha y\rVert = 1$

However, by metric and norm properties,

$$\lVert\alpha x-\alpha y\rVert =\lVert\alpha(x-y)\rVert = \lvert\alpha\rvert\,\lVert x-y\rVert\le\vert \alpha \vert.1=\alpha,$$ a contradiction!!

Styles
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Say we have a normed field $K$ and that $X$ is a normed space over $K$.

The question has positive answer (i.e. discrete topology is not induced by any norm) if we implicitely assume that the normed field has a non-trivial norm. But if we further investigate into pathological cases, we can in fact find a normed space in which a trivial norm makes sense. So we have a counterexample.

Say $K$ is a finite field, then it can be shown that the only norm for $K$ is the trivial one. Now if we take the $K$-vector space $X = K^n$, then by fixing a basis (say, the canonical basis), there is a canonical way to define a norm over $X$, namely $$ |(x_1, \dots, x_n)| := \max_{i = 1, \dots, n} |x_i|, $$ and it turns out to be a trivial norm over $X$. So the induced topology is the discrete topology.

So, to be fair, we can say that the trivial topology is not induced by any norm if we assume that $K$ has a non-trivial topology (the argument is the same given by Brian M. Scott and by user38268).

fmnq
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