# Difference between metric and norm made concrete: The case of Euclid

This is a follow-up question on this one. The answers to my questions made things a lot clearer to me (Thank you for that!), yet there is some point that still bothers me.

This time I am making things more concrete: I am esp. interested in the difference between a metric and a norm. I understand that the metric gives the distance between two points as a real number. The norm gives the length of a a vector as a real number (see def. e.g. here). I further understand that all normed spaces are metric spaces (for a norm induces a metric) but not the other way around (please correct me if I am wrong).

Here I am only talking about vector spaces. As an example lets talk about Euclidean distance and Euclidean norm. Wikipedia says:

A vector can be described as a directed line segment from the origin of the Euclidean space (vector tail), to a point in that space (vector tip). If we consider that its length is actually the distance from its tail to its tip, it becomes clear that the Euclidean norm of a vector is just a special case of Euclidean distance: the Euclidean distance between its tail and its tip.

What confuses me is that they seem to be having it backwards: The Euclidean metric induces the Euclidean norm: You measure the distance between tip and tail and get the length out of that. What makes my confusion complete is that $L^2$ distance is also called the Euclidean norm (see here).

I would very much appreciate it if somebody could clear the haze.

• you are correct that all normed spaces are metric spaces. Just define $d(x,y) = \|x - y \|$ – Juan S May 12 '11 at 7:03
• I am confused about what you're confused about. You can define the norm in terms of the metric, or you can define the metric in terms of the norm. It doesn't matter which order you do it in. – Qiaochu Yuan May 12 '11 at 7:12
• @Qiaochu: Thank you. Perhaps my confusion stems from the fact that on the one hand, as you say, you can do it either way, on the other hand there is this hierarchy between metric spaces and normed spaces (every normed space is a metric space but not the other way round). It seems like a contradiction to me...?!? – vonjd May 12 '11 at 7:42
• given a norm $|| \cdot ||$, you can always define a metric $d(x, y) = ||x - y||$ as Qwirk says. Given a metric $d(x, y)$ on a vector space, you can define a function $d(x, 0)$, and sometimes it's a norm and sometimes it isn't. There is no contradiction here. – Qiaochu Yuan May 12 '11 at 8:10
• I think I finally got it - Thank you to everybody! – vonjd May 12 '11 at 10:52

The metric $d(u,v)$ induced by a vector space norm has additional properties that are not true of general metrics. These are:

Translation Invariance: $d(u+w,v+w)=d(u,v)$

Scaling Property: For any real number $t$, $d(tu,tv)=|t|d(u,v)$.

Conversely, if a metric has the above properties, then $d(u,0)$ is a norm.

More informally, the metric induced by a norm "plays nicely" with the vector space structure. The usual metric on $\mathbb{R}^n$ has the two properties mentioned above. But there are metrics on $\mathbb{R}^n$ that are topologically equivalent to the usual metric, but not translation invariant, and so are not induced by a norm.

• @user6312: Thank you - could you please give an example of a simple metric that is not translation invariant? – vonjd May 12 '11 at 7:37
• @vonjd: on $\mathbb{R}$, let $d(x, y) = 0$ if $x = y$, and otherwise let $d(x, y) = |x| + |y|$. Intuitively $d(x, y)$ measures the length of the shortest path from $x$ to $y$ if whenever $x \neq y$ such a path must travel to the origin first. – Qiaochu Yuan May 12 '11 at 8:14
• @vonjd: Another example: $d(x,y)=\sqrt{\lvert x-y\rvert}$. This generalises to composing any metric with a concave monotonous function $[0,\infty)\to [0,\infty)$ mapping $0$ to $0$. – Rasmus May 12 '11 at 11:35

The simplest answer to the question in the title is that a metric is a function of two variables and a norm is a function of one variable.

The other question, which I would summarize as "which came first?" is (at least in the Euclidean context) a chicken-and-egg question. You can define the Euclidean (or $L^2$) distance between $x$ and $y$ as $\sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}$ and then define the norm as the distance from $x$ to the origin; or, you can define the Euclidean (or $L^2$) norm as $\sqrt{x_1^2+\cdots+x_n^2}$ and then define the distance from $x$ to $y$ as the norm of $x-y$.