Simple example of a metric space that is not a normed space

Key takeaway: A metric space only needs a distance function. A normed space is much more structured: the metric has to come from a norm on a vector space. So the easiest counterexamples are finite sets or spaces where no linear structure is even present.

A very simple example is the set $X=\{a,b\}$ with the metric $d(a,b)=1,\quad d(a,a)=0,\quad d(b,b)=0.$ This is a metric space, but it is not a normed space because there is no vector space structure on $X$ at all.

If you want a more familiar example, you can also use any set with the discrete metric: $d(x,y)=\begin{cases}0,&x=y\\1,&x\ne y\end{cases}$ That is definitely a metric space. But unless the set has a suitable vector space structure and the metric comes from a norm, it is not a normed space.

So the shortest counterexample is:

• a two-point set with the discrete metric.

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Pitfalls the pros know 👇 The usual trap is to think every metric space automatically has addition and scalar multiplication. It does not. Normed spaces are always vector spaces first, then the norm creates the metric. If the set has no vector space structure, it cannot be a normed space.

What if the problem changes? If you want an example on an infinite set, the integers $\mathbb{Z}$ with the discrete metric also work as a metric space. The same issue remains: without a vector space structure over a field, it is not a normed space.

Tags: discrete metric, vector space, normed space