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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.

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    Except if $X=\mathbb F_2$ the vector space over the field $\mathbb F_2$ with two elements.2012-09-21

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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 $X$.

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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.2012-09-21
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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 $\dfrac{\epsilon}{2\alpha}y$?

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    @Ben: What’s obvious depends so much on one’s background. A $f$unctional analyst would $p$robably agree with you; I’m a set-theoretic topologist, so it’s not really surprising that I $f$ind the topological solution much more obvious.2012-09-21
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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).