# Inducing metric of a vector space

What does it mean, in the context of normed vector spaces, that the norm induces the metric? Furthermore, why normal vector spaces can't have a metric and be considered a metric space then?

You have a norm on your vector space. It tells you more than just the distance between points. So it induces a metric because it tells you what the metric is, i.e. $$d(x,y)=||x-y||$$, but a norm is more than just a metric.

Un-normed vector spaces can be metric, but it's classic result that there is only one vector space topology on a finite dimensional vector space over $$\mathbb{R}$$, and this topology is necessarily normed.

In infinite dimension, I guess the most classic example is $$C^{\infty}(K)$$ for some compact $$K\subseteq \mathbb{R}$$ with the metric given by

$$d(f,g)=\sum_{k=0}^{\infty} 2^{-k} \min\{1,||f^{(k)}-g^{(k)}||_{\infty}\} ,$$

which isn't equivalent to any norm.

• small formatting comment: $||x||$ produces $||x||$ and $\|x\|$ produces $\|x\|$
– Surb
Oct 6 '19 at 12:00
• So the difference in output is the spacing? Oct 6 '19 at 12:02
• Yes, excatly. Just compare these to expressions: $\||f|\|$ and $|||f|||$. Oct 6 '19 at 12:14
• Actually the example of $C^\infty$ is quite nice for OPs last question, because $C^\infty(\mathbb{R})$ is not metrizable! Oct 6 '19 at 12:17
• What topology are you giving it? Naturally, $f_n\to f$ in $C^{\infty}(\mathbb{R})$ if $f_n\to f$ in $C^{\infty}([-n,n])$ for every $n$, but this is convergence in countably many metrics, and hence, metric convergence. $C^{\infty}_c(\mathbb{R})$, however, isn't metric when given its usual topology. Oct 6 '19 at 12:20

If you have a normed vector space $$(V,\|\cdot\|)$$, simply by setting $$d(u,v) := \|u-v\|$$ for all $$u,v \in V$$ you can define and thus induce a metric on $$V$$. Of course there are other choices.

Now, a vector space is more an algebraic object because it combines fields and groups. Yet, you can always define at least a topology on those "normal Vector spaces" so that at least the scalar multiplication and the addition is continuous (called topological vector spaces). Many of these vector spaces you are familiar with also will allow for more structur, e.g. a metric or a norm. But there are (topological) vector spaces that for some topological reasons have to "many" open sets that it can been shown that those vector spaces are not metrizable in a nice and useful way.

That may sound cryptic but one key point here is the dimension of the given vector space. If it is finite, you are fine and actually every norm you can find is even equivalent. But when it comes to infinite dimensional vector spaces, things get vastly more interesting. Now, you have to be clear about one thing, when it comes to normed or metric spaces, the most important feature is the metric or norm. You can have the exact same set or vector spaces but endowe it with different norms and you get from a certain perspective very different objects.