Isometries

Maps That Preserve Every Distance

A Lipschitz map controls how much distances can expand. An isometry imposes the sharper condition that every distance is preserved exactly. This gives a strong comparison between the geometry of the domain and the geometry of the map's image, even when the map does not cover its entire target.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. The equality in the definition below is required for every pair of points, not merely for nearby points. It immediately gives a Lipschitz bound, but it also prevents the map from collapsing distinct points together.

Definition (Isometric Embedding): A map $f:X\to Y$ is an isometric embedding if $$ d_Y(f(x),f(y))=d_X(x,y) $$ for all $x,y\in X$. A surjective isometric embedding is called a surjective isometry. In this tutorial, “isometry” will mean a surjective isometric embedding; some texts use the word more broadly for any distance-preserving map.

Every isometric embedding is injective. Indeed, if $f(x)=f(y)$, then $d_X(x,y)=d_Y(f(x),f(y))=0$, so the metric property gives $x=y$. Thus an isometric embedding identifies $X$ with its image $f(X)$, equipped with the metric restricted from $Y$. It need not be onto $Y$.

An isometric embedding is Lipschitz with constant $1$, by the defining equality. If $X$ has at least two points, every ratio of output distance to input distance is exactly $1$, so its best Lipschitz constant is $1$, using the Best Lipschitz Constant Theorem from “Lipschitz Maps.” For a space with at most one point, the distance-preserving condition is vacuous or involves only the pair $x=x$; the least Lipschitz constant is then $0$.

Distances, Balls, and Subsets

Exact preservation of distances gives an exact description of how balls intersect the image. Here $B_r(x)$ denotes a ball in $X$, while a ball centered at $f(x)$ without further qualification is taken in $Y$.

Theorem (Balls Under an Isometric Embedding): If $f:X\to Y$ is an isometric embedding, $x\in X$, and $r>0$, then $$ f(B_r(x))=f(X)\cap B_r(f(x)). $$

Proof. Suppose first that $z\in f(B_r(x))$. Then $z=f(u)$ for some $u\in X$ with $d_X(u,x)<r$. Since $f$ preserves distances, $$ d_Y(z,f(x))=d_Y(f(u),f(x))=d_X(u,x)<r. $$ Also $z\in f(X)$, so $z\in f(X)\cap B_r(f(x))$.

Conversely, suppose $z\in f(X)\cap B_r(f(x))$. There is $u\in X$ with $z=f(u)$, and $d_Y(f(u),f(x))<r$. Distance preservation gives $d_X(u,x)<r$, so $u\in B_r(x)$ and $z\in f(B_r(x))$. Both inclusions hold, proving the equality. $\square$

The intersection with $f(X)$ is essential. The ball $B_r(f(x))$ may contain points of $Y$ that are not in the image at all. The theorem says that the points of the target ball that do belong to the image correspond exactly to the original ball in $X$.

The same reasoning applies to distance from a point to a nonempty subset. For a nonempty set $E$ in a metric space, write $d(x,E)=\inf\{d(x,e):e\in E\}$, as in “Closure in Metric Spaces.”

Theorem (Distances to Subsets Under an Isometric Embedding): Let $f:X\to Y$ be an isometric embedding, let $E\subseteq X$ be nonempty, and let $x\in X$. Then $$ d_Y(f(x),f(E))=d_X(x,E). $$

Proof. The set $f(E)$ is nonempty because $E$ is nonempty. As $e$ ranges over $E$, the defining equality for an isometric embedding gives $$ \{d_Y(f(x),z):z\in f(E)\} =\{d_Y(f(x),f(e)):e\in E\} =\{d_X(x,e):e\in E\}. $$ The two sets of distances are identical, so their infima are equal. This is exactly the asserted formula. $\square$

Worked Example: A Diagonal Copy of the Real Line

Give $\mathbb{R}$ its usual metric and $\mathbb{R}^2$ the Euclidean metric. Define $f:\mathbb{R}\to\mathbb{R}^2$ by $f(x)=(x/\sqrt{2},x/\sqrt{2})$. For $x,y\in\mathbb{R}$, direct calculation gives

d_2(f(x),f(y)) =\sqrt{\left(\frac{x-y}{\sqrt{2}}\right)^2+\left(\frac{x-y}{\sqrt{2}}\right)^2} =\sqrt{\frac{(x-y)^2}{2}+\frac{(x-y)^2}{2}} =|x-y|.

Thus $f$ is an isometric embedding. Its image is the diagonal line $\{(u,u):u\in\mathbb{R}\}$, which is not all of $\mathbb{R}^2$, so this embedding is not a surjective isometry onto the plane. To check that its image is closed, suppose a sequence $(u_n,u_n)$ in the image converges in $\mathbb{R}^2$ to $(a,b)$. Coordinate convergence gives $u_n\to a$ and $u_n\to b$. Uniqueness of metric limits implies $a=b$, so the limit $(a,b)=(a,a)$ is still on the diagonal.

Worked Example: An Isometric Embedding with Nonclosed Image

Consider the inclusion $f:(0,1)\to[0,1]$, with the usual metric on each space. For $x,y\in(0,1)$, $d(f(x),f(y))=|x-y|=d(x,y)$, so $f$ is an isometric embedding. It is not surjective, since neither $0$ nor $1$ belongs to its image.

Its image is also not closed in $[0,1]$. For each integer $n\geq1$, let $x_n=1/(n+1)$. Then $x_n\in(0,1)$, and $x_n\to0$ in $[0,1]$, but $0\notin f((0,1))$. The Sequential Characterization of Closed Sets therefore shows that the image is not closed. This example also illustrates the role of completeness: $(0,1)$ with its usual metric is not complete, since $(x_n)$ is Cauchy in $(0,1)$ but has no limit there.

Worked Example: The Same Real Line with a Different Metric

On $Y=(0,\infty)$, define $\rho(u,v)=|\ln u-\ln v|$. This is a metric: it is the metric induced by the injective function $\ln:(0,\infty)\to\mathbb{R}$, as in “Examples of Metrics.” Define $f:\mathbb{R}\to Y$ by $f(x)=e^x$. Since $\ln(e^x)=x$, for all $x,y\in\mathbb{R}$,

\rho(f(x),f(y)) =|\ln(e^x)-\ln(e^y)| =|x-y|.

So $f$ is distance-preserving. It is also surjective: if $u>0$, then $u=e^{\ln u}=f(\ln u)$. Hence $f$ is a surjective isometry between these metric spaces, although the formula $x\mapsto e^x$ would not preserve distances if the target used the usual metric $|u-v|$. Which maps are isometries depends on the metrics, not just on the underlying sets or formulas.

Complete Domains Have Closed Images

The nonclosed image in the preceding example is consistent with a useful general fact: an isometric embedding of a complete space has a closed image, even if the target space is not complete. Completeness prevents a sequence in the image from converging to a missing point of the target.

Theorem (Complete Domains Have Closed Isometric Images): Let $X$ be a complete metric space, let $Y$ be any metric space, and let $f:X\to Y$ be an isometric embedding. Then $f(X)$ is closed in $Y$.

Proof. Let $(y_n)$ be a sequence in $f(X)$ that converges in $Y$ to some $y\in Y$. For each $n$, choose $x_n\in X$ with $f(x_n)=y_n$. We show that $(x_n)$ is Cauchy. Given $\varepsilon>0$, convergence of $y_n$ to $y$ gives an index $N$ such that, for $n,m\geq N$, $$ d_Y(y_n,y_m)\leq d_Y(y_n,y)+d_Y(y,y_m)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ Distance preservation then yields $d_X(x_n,x_m)=d_Y(f(x_n),f(x_m))=d_Y(y_n,y_m)<\varepsilon$. Thus $(x_n)$ is Cauchy in $X$.

By completeness, there is $x\in X$ such that $x_n\to x$. Again using distance preservation, $$ d_Y(y_n,f(x))=d_Y(f(x_n),f(x))=d_X(x_n,x)\longrightarrow0. $$ Therefore $y_n\to f(x)$ in $Y$. The sequence also converges to $y$, so uniqueness of metric limits gives $y=f(x)\in f(X)$. We have shown that every convergent sequence in $f(X)$ has its limit in $f(X)$. The Sequential Characterization of Closed Sets now implies that $f(X)$ is closed in $Y$. $\square$

The target need not be complete in this theorem. The proof uses completeness only for the sequence $(x_n)$ in the domain. The isometry transfers the Cauchy behavior of the image sequence back to $X$, and then transfers the domain limit forward again.

In the other direction, care is needed: a closed image alone does not force the domain to be complete if the target is not complete. For instance, the identity map from $\mathbb{Q}$ to $\mathbb{Q}$, with the usual metric, is an isometric embedding with closed image in $\mathbb{Q}$, but $\mathbb{Q}$ is incomplete. When the target is complete, the familiar Closed Subset of a Complete Space Theorem from “Complete Metric Spaces” gives a converse: if an isometric embedding has closed image in a complete target, then its image is complete, and distance preservation makes the domain complete as well.

What Is Preserved—and What Is Not

An isometric embedding preserves the distance between every pair of domain points, and therefore preserves more than just a bound on changes. It preserves distances to subsets of the form $f(E)$, as proved above, and it carries balls to the corresponding balls relative to its image. In particular, it preserves the metric information within $X$.

However, an isometric embedding does not assert that every target point belongs to the image, nor that the image is closed. The inclusion $(0,1)\to[0,1]$ demonstrates both failures. To conclude that the image is closed, a sufficient hypothesis is completeness of the domain. To conclude that a distance-preserving map is a surjective isometry, surjectivity must be established separately.

Check exact preservation.
For every pair $x,y$, verify $d_Y(f(x),f(y))=d_X(x,y)$, using the stated metrics on both spaces.

Separate embedding from surjectivity.
Distance preservation implies injectivity, but it does not imply that every point of $Y$ is attained.

Check closedness with the right hypothesis.
A complete domain guarantees a closed image. Without completeness, an isometric embedding can have a nonclosed image.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Why must every isometric embedding be injective?
Why does the formula for balls involve intersecting the target ball with $f(X)$?
If $f:X\to Y$ is an isometric embedding and $E\subseteq X$ is nonempty, what is the relation between $d_X(x,E)$ and $d_Y(f(x),f(E))$?
What sequence demonstrates that the inclusion $(0,1)\to[0,1]$ has a nonclosed image?
Why does completeness of $X$ imply that the image of an isometric embedding $f:X\to Y$ is closed, even when $Y$ is not complete?

What You'll Learn