# Show that the discrete metric can not be obtained from $X\neq\{0\}$

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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The hypothesis that $X\ne\{0\}$ doesn’t mean that $0$ is not in $X$: it means that $X$ is not the trivial one-point vector space. – Brian M. Scott Sep 21 '12 at 0:32
Except if $X=\mathbb F_2$ the vector space over the field $\mathbb F_2$ with two elements. – Did Sep 21 '12 at 15:07

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. – Ben Millwood Sep 21 '12 at 13:50

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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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 1:29
@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 6:42
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