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(If this question may expose some fundamental misunderstanding about normed linear spaces and metric spaces, I appreciate any corrections to these misconceptions in lieu of answering the question directly.)

I understand that if we have a normed linear space $(X, \|\cdot\|)$, it is also a metric space with the metric $d(x,y)=\|x-y\|$. Is it reasonable/possible/sensible to define a metric space from $X$ without using the norm as the metric? e.g. could it make sense to define a metric space from the normed linear space by equipping it with the discrete metric instead?

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It makes sense to define the discrete metric on any set. But why would you do that on a real vector space $V$? Then, for instance, if you take $v\in V\setminus\{0\}$, the map$$\begin{array}{ccc}\mathbb R&\longrightarrow&V\\\lambda&\mapsto&\lambda v\end{array}$$will not be continuous.

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  • $\begingroup$ I suppose my question should have been, "why do we say that 'a normed linear space is a metric space' with the specific metric $d(x,y)=||x-y||$?" It seems that we're arbitrarily creating a metric on the original vector space (we just happen to be choosing one that's very reasonable). So, if we're taking a normed vector space and assigning an arbitrary metric to it to make it a metric space, what makes it special? E.g. why can't we extend this same argument to conclude that 'any set is a metric space?' $\endgroup$
    – dkv
    Oct 2 '19 at 18:26
  • $\begingroup$ On any normed vector space you define, as you wrote, $d(x,y)=\lVert x-y\rVert$. How do you do the same thing on an arbitrary set? $\endgroup$ Oct 2 '19 at 20:08
  • $\begingroup$ The "argument" would be that since we're defining $d(x,y)=||x-y||$ arbitrarily, we can simply pick another metric e.g. $d(x,y) = 1$ if $x=y$ and $0$ otherwise. This is a weird choice to make, but I can't see the formal reason we aren't allowed to do it. Continuing with this reasoning, since we're imposing the discrete metric $d(x,y)$ on the normed linear space, we can similarly impose the discrete metric onto any set and conclude it's a metric space. $\endgroup$
    – dkv
    Oct 2 '19 at 21:04
  • $\begingroup$ Yes, we can do that. But then, when we are dealing with a vector space, we get a distance that has no relation whatsoever with the vector space structure. And I must say that I think that it is rather strange that you find arbitrary the choice of taking $d(x,y)=\lVert x-y\rVert$. The norm is already there and the map $(x,y)\mapsto\lVert x-y\rVert$ is a distance! $\endgroup$ Oct 2 '19 at 21:39

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