The first problem in multivariable analysis is that points in $\mathbb{R}^n$ are not arranged on a line. In $\mathbb{R}$, the distance between $x$ and $y$ is $|x-y|$, and limits, continuity, open intervals, derivatives, and compactness all lean on that single quantity. In $\mathbb{R}^n$, a displacement has several components, so we need a way to turn the coordinate differences into one nonnegative number that behaves like distance.

The Euclidean metric is the standard answer. It measures the length of the straight-line displacement from $x$ to $y$ by taking the square root of the sum of the squared coordinate differences. This is the metric secretly used whenever we draw circles in the plane, balls in space, or small neighbourhoods in a coordinate chart.

A bad distance formula breaks analysis quickly. If we tried to measure displacement in $\mathbb{R}^2$ by adding coordinates with signs, the distance from $(0,0)$ to $(1,-1)$ would be $0$, even though the points are different. If we ignored one coordinate, sequences could move far away in the ignored direction while appearing to converge. The Euclidean metric avoids these failures by measuring every coordinate symmetrically and by making the [Pythagorean theorem](/theorems/3266) part of the geometry.

[example: A Failed Coordinate Sum] Define $s:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ by \begin{align*} s(x,y)=(x_1-y_1)+(x_2-y_2) \end{align*} for $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Take $x=(0,0)$ and $y=(1,-1)$. These points are distinct because their first coordinates are different: $0\ne1$. But the coordinate-sum formula gives \begin{align*} s((0,0),(1,-1))=(0-1)+(0-(-1)) \end{align*} and hence \begin{align*} s((0,0),(1,-1))=-1+1=0. \end{align*} Therefore $s$ assigns value $0$ to two distinct points, so it cannot satisfy the separation requirement for a metric. Replacing $s$ by $|s|$ does not repair the failure, because \begin{align*} |s((0,0),(1,-1))|=|0|=0. \end{align*} The problem is cancellation: the displacement from $x$ to $y$ is \begin{align*} x-y=(0-1,0-(-1))=(-1,1), \end{align*} and the signed aggregate $(-1)+1$ loses the fact that both coordinates changed. A genuine distance must control the whole displacement vector rather than only the sum of its signed coordinate differences. [/example]

The rest of the chapter builds the Euclidean metric from the Euclidean norm, checks that it satisfies the metric axioms, and explains why this single formula controls the familiar topology, convergence, compactness, and calculus of $\mathbb{R}^n$.

## Definition

The page topic is the distance function itself, so we define it before abstracting its ingredients. The formula should assign zero distance only to equal points, should be unchanged when the two points are swapped, and should measure all coordinate differences without allowing positive and negative coordinates to cancel. Squaring the coordinate differences is the simplest way to enforce that behaviour while preserving the Pythagorean geometry of straight-line displacement.

[definition: Euclidean Metric] For $n\in\mathbb{N}$, the Euclidean metric on $\mathbb{R}^n$ is the function $d_2:\mathbb{R}^n\times\mathbb{R}^n\to[0,\infty)$ defined by \begin{align*} d_2(x,y) = \left(\sum_{k=1}^n |x_k-y_k|^2\right)^{1/2} \end{align*} for $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$. [/definition]

The notation $d_2$ records that the formula uses squares and a square root. In most analysis on $\mathbb{R}^n$, this formula is so standard that authors write $|x-y|$ instead of $d_2(x,y)$. This shorthand only works because there is already a notion of the length of a single displacement vector. Before using $|x-y|$ freely, we need to name that length function and distinguish it from the two-point distance it generates.

[definition: Euclidean Norm] For $n\in\mathbb{N}$, the Euclidean norm is the function $|\cdot|:\mathbb{R}^n\to[0,\infty)$ defined by \begin{align*} |x| = \left(\sum_{k=1}^n |x_k|^2\right)^{1/2} \end{align*} for $x=(x_1,\dots,x_n)$. [/definition]

The square root is not decoration: it makes $|x|^2$ additive across perpendicular coordinate directions. That is the algebraic form of the Pythagorean theorem, and it is why spheres, rotations, and orthogonal projections fit naturally with this norm. The norm gives lengths of vectors, while the Euclidean metric applies that length formula to the displacement vector $x-y$. The problem now is certification: before this formula can support limits and open sets, we need an abstract checklist for what any distance function must satisfy. That checklist is the metric definition.

[definition: Metric] Let $X$ be a set. A metric on $X$ is a function $d:X\times X\to[0,\infty)$ such that for all $x,y,z\in X$, \begin{align*} d(x,y)=0 \iff x=y, \end{align*} \begin{align*} d(x,y)=d(y,x), \end{align*} and \begin{align*} d(x,z)\le d(x,y)+d(y,z). \end{align*} [/definition]

The first condition says distance separates points. The second says distance has no direction. The third, the triangle inequality, says a direct path is no longer than a path through an intermediate point. For the Euclidean formula, the separation and symmetry properties follow directly from the coordinate expression; the serious estimate is the triangle inequality.

For vectors $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ in $\mathbb{R}^n$, write

\begin{align*} x\cdot y=\sum_{k=1}^n x_k y_k \end{align*}

for the Euclidean dot product, and write $|x|=\sqrt{x\cdot x}$ for Euclidean length. The estimate needed here is the coordinate Cauchy--Schwarz inequality:

\begin{align*} |x\cdot y|\le |x|\,|y|. \end{align*}

It says that the dot product cannot exceed the product of the two lengths. This elementary form is enough to prove $|x+y|\le |x|+|y|$, so the coordinate formula becomes a genuine metric rather than just a plausible distance candidate.

[quotetheorem:8512]

This theorem is the permission slip for using every construction from [metric spaces](/page/Metric%20Space) in $\mathbb{R}^n$. Once $d_2$ is known to be a metric, open balls, closed sets, limits, continuity, and compactness can be defined without returning to coordinates every time.