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.
 A: 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 $\dfrac{\epsilon}{2\alpha}y$?
A: 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 $X$.
A: 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!!
A: 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).
