Real Analysis: Metric Spaces
Isometries
Isometries: Preserving Distances Between Points
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 $a, b \in X.$ In this case, $X$ and $Y$ are said to be isometric. If an isometry is bijective, it is called a global isometry, and the two spaces in question are globally isometric. 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.
Preservation of Topological Properties
Subsequent sections will cover the topological properties of metric spaces and their points and subsets. Because metric topology is built on the notion of distance in metric spaces, the topological properties of sets and spaces are preserved under global isometries. These properties are listed here for ease of reference.
If $f : (X, d_X) \rightarrow (Y, d_Y)$ is a global isometry, then
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$S$ is open in $(X, d_X)$ if and only if $f(S)$ is open in $(Y, d_Y).$
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$S$ is closed in $(X, d_X)$ if and only if $f(S)$ is closed in $(Y, d_Y).$
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$S$ is bounded in $(X, d_X)$ if and only if $f(S)$ is bounded in $(Y, d_Y).$
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The interior of $S$ is preserved: $f(S^{\circ}) = f(S)^{\circ}.$
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The boundary of $S$ is preserved: $f(\partial S) = \partial f(S).$
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The closure of $S$ is preserved: $f(\overline{S}) = \overline{f(S)}.$
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The exterior of $S$ is preserved: $f(\text{Ext}(S)) = \text{Ext}(f(S)).$
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$x$ is an isolated point of $S$ if and only if $f(x)$ is an isolated point of $f(S).$
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$x$ is a limit point of $S$ if and only if $f(s)$ is an isolated point of $f(S).$
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$(X, d_X)$ is connected if and only if $(Y, d_Y)$ is connected.
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$(X, d_X)$ is complete if and only if $(Y, d_Y)$ is complete.
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$S$ is compact in $(X, d_X)$ if and only if $f(S)$ is compact in $(Y, d_Y).$
Open and closed sets are the foundation of the concepts of connectedness, completeness, and compactness. Notably, non-global isometries are not guaranteed to preserve open or closed sets, and therefore are not guaranteed to preserve the other topological concepts either.
Problems
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}) $
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}) $
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.
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.
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.
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.