# Real Analysis: Metric Spaces

## Isometries

An isometry is a function between two metric spaces that preserves the distance between points. Formally, given two metric spaces $(X, d_X)$ and $(Y, d_Y)$, an isometry is a function $f : X \rightarrow Y$ where $d_X(a, b) = d_Y(f(a), f(b))$ for all $a, b \in X.$ If an isometry is bijective, it is called a global isometry. The word "isometry" comes from the Greek prefix "isos," which means "equal," and "metron," which means "measure."

In Euclidean space, isometries are rigid body movements. That is to say, the image of a shape under an isometry will be the same as the preimage, just oriented differently in space, rather than stretched, sheared, broken, or otherwise deformed. The three main kinds of isometry are translation, rotation, and reflection.

## Problems

1. Translation: Consider the function $f$ from $(\mathbb{R}^n, d)$ to itself, where $d(\textbf{x}, \textbf{y}) = \|\textbf{x}-\textbf{y}\|_p$, where $f(\textbf{x}) = \textbf{x} + \textbf{c}$, and where $\textbf{c}$ is any constant. Show that $f$ is an isometry.

$d(f(\textbf{x}), f(\textbf{y})) = d(\textbf{x} + \textbf{c}, \textbf{y} + \textbf{c}) \\ d(f(\textbf{x}), f(\textbf{y})) = \sqrt[p]{\sum\limits_{i=1}^n ((x_i + c_i) - (y_i + c_i))^p} \\ d(f(\textbf{x}), f(\textbf{y})) = \sqrt[p]{\sum\limits_{i=1}^n (x_i - y_i)^p} \\ d(f(\textbf{x}), f(\textbf{y})) = d(\textbf{x}, \textbf{y})$

2. Reflection: Consider the function $f$ from $(\mathbb{R}^n, d)$ to itself where $d(\textbf{x}, \textbf{y}) = \|\textbf{x} - \textbf{y}\|_p$ and $f(\textbf{x}) = -\textbf{x}$. Show that $f$ is an isometry.

$d(f(\textbf{x}), f(\textbf{y})) = d(-\textbf{x}, -\textbf{y}) \\ d(f(\textbf{x}), f(\textbf{y})) = \sqrt[p]{\sum\limits_{i=1}^n ((-x_i) - (-y_i))^p} \\ d(f(\textbf{x}), f(\textbf{y})) = \sqrt[p]{\sum\limits_{i=1}^n (y_i - x_i)^p} \\ d(f(\textbf{x}), f(\textbf{y})) = d(\textbf{y}, \textbf{x}) \\ d(f(\textbf{x}), f(\textbf{y})) = d(\textbf{x}, \textbf{y})$

3. Let $f$ be a function from $X$ to itself under the discrete metric, $d$. Show that $f$ is an isometry if and only if it is injective.

Assume $f$ is injective. If $f(x) = f(y)$, then $x = y$, so $d(f(x), f(y)) = 0 = d(x,y)$. If instead $f(x) \neq f(y)$, then $x \neq y$, so $d(f(x), f(y)) = 1 = d(x,y)$. Therefore $f$ is an isometry.

Conversely, assume $f$ is an isometry. Then $d(x, y) = d(f(x), f(y))$. If $f(x) = f(y)$, then $d(f(x), f(y)) = 0 = d(x, y)$, so $x = y$. Therefore $f$ is injective.

4. Show that every isometry is injective.

Let $f$ be an isometry from $(X, d_X)$ to $(Y, d_Y)$. Then $d_X(x, y) = d_Y(f(x), f(y))$. If $f(x) = f(y)$, then $d_Y(f(x), f(y)) = 0$, so $d_X(x, y) = 0$, and thus $x = y$. Therefore $f$ is injective.

5. Given an example of an injective function that is not an isometry.

Consider $f : \mathbb{R} \rightarrow \mathbb{R}$ where $f(x) = 2x.$ If $f(a) = f(b),$ then $2a = 2b,$ and thus $a = b,$ and so $f$ is injective. However, $d(f(5), f(6)) = |10 - 12| = 2,$ but $d(5, 6) = |5 - 6| = 1,$ and so $f$ is not an isometry.

6. Let $f$ and be an isometry from $(X, d_X)$ into $(Y, d_Y),$ and let $g$ be an isometry from $(Y, d_Y)$ into $(Z, d_Z).$ Show that the composition $g \circ f$ is also an isometry.

Since $f$ is an isometry, it follows that $d_X(a, b) = d_Y(f(a), f(b)).$ Likewise, since $g$ is an isometry, it follows that $d_Y(x, y) = d_Z(g(x), g(y)).$ Set $x = f(a)$ and $y = f(b).$ Then $d_X(a, b) = d_Y(f(a), f(b)) = d_Z(g(f(a)), g(f(b)).$ Therefore $g \circ f$ is an isometry.