Every Normed Vector Space is a Metric Space

A normed vector space has more structure than a metric space. It has vector addition, scalar multiplication, and a norm. The metric space only needs a way to measure distance.

The bridge is subtraction. In a vector space, two points $x$ and $y$ determine a displacement $x-y$. The norm measures the length of that displacement. So the distance between $x$ and $y$ is not added as new information; it is extracted from the norm:

$$d(x,y)=\Vert x-y\Vert .$$

The metric axioms are exactly the norm axioms seen through this displacement.

The distance is non-negative because norms are non-negative:

$$d(x,y)=\Vert x-y\Vert \ge 0.$$

It is zero exactly when the two points are the same:

$$d(x,y)=0⟺\Vert x-y\Vert =0⟺x-y=0⟺x=y.$$

It is symmetric because reversing the displacement only multiplies it by $-1$:

$$d(y,x)=\Vert y-x\Vert =\Vert -(x-y)\Vert =|-1|\Vert x-y\Vert =\Vert x-y\Vert =d(x,y).$$

It satisfies the triangle inequality because the direct displacement decomposes through an intermediate point $z$:

$$x-y=(x-z)+(z-y),$$

so

$$d(x,y)=\Vert x-y\Vert =\Vert (x-z)+(z-y)\Vert \le \Vert x-z\Vert +\Vert z-y\Vert =d(x,z)+d(z,y).$$

The important point is that a normed vector space becomes a metric space by forgetting structure. The metric remembers distances, but it forgets how to add points, scale vectors, and talk about linear combinations.

So every normed vector space is a metric space, but not every metric space is a normed vector space. A metric space may have distances without any meaningful notion of displacement.