# Understanding isometric spaces

I have studied that an isometry is a distance-preserving map between metric spaces and two metric spaces $X$ and $Y$ are called isometric if there is a bijective isometry from X to Y.

My questions are related with the understanding of isometric spaces, they are as follows:

Can we say that two isometric spaces are same? If no, in what context they differ? What are the common properties shared by two isometric spaces?

Intuitively what are isometric spaces?

If two spaces are isometric how to find out bijective distance preserving map between them?

Thanks for your help and time.

-

1
Any two lines in the plane are isometric (you can translate and rotate to place one line on top of the other, without affecting distances within the lines in this process), so definitely two isometric spaces need not really be literally the same. But as far as metric properties are concerned they behave in the same way. It's like asking "are all circles of radius 1 the same"? No, but obviously you're comfortable using one particular choice (like the one centered at the origin) even if that's not the original circle of interest. – KCd May 28 '12 at 6:04

1
The examples of the line and circle might seem silly, because they are very familiar. The point of the concept of isometric spaces is to keep us aware that we shouldn't consider two isometric spaces as being fundamentally different from one another. For example, when you construct the completion $\widetilde{X}$ of a metric space $X$, using equiv. classes of Cauchy sequences in $X$, you don't really find $X$ as a subset of $\widetilde{X}$, but $X$ is isometric to the equiv. classes of constant seq. $(x,x,x,\dots)$, and that is how we can view $X$ (as metric space) inside $\widetilde{X}$. – KCd May 28 '12 at 6:08

1
There is no universal method to find an isometry between two isometric metric spaces. – KCd May 28 '12 at 6:09
@KCd Thanks to you. Your comments are helpful to me. – srijan May 28 '12 at 6:17

## 1 Answer

Homeomorphisms are the maps that preserve all topological properties: from a structural point of view, homeomorphic spaces might as well be identical, though they may have very different underlying sets, and if they’re metrizable, they may carry very different (but equivalent) metrics. Isometries are the analogue for metric spaces, topological spaces carrying a specific metric: they preserve all metric properties, and of course those include the topological properties. Thus, all isometries are homeomorphisms, but the converse is false.

Consider the metric spaces $\langle X,d_X\rangle$ and $\langle Y,d_Y\rangle$ defined as follows: $X=\Bbb N,Y=\Bbb Z$, $$d_X(m,n)=\begin{cases}0,&\text{if }m=n\\1,&\text{if }m\ne n\;,\end{cases}$$ for all $m,n\in X$, and $$d_Y(m,n)=\begin{cases}0,&\text{if }m=n\\1,&\text{if }m\ne n\end{cases}$$ for all $m,n\in Y$. It’s easy to check that $d_X$ and $d_Y$ are metrics on $X$ and $Y$, respectively.

Clearly these are not the same space: they have different underlying sets. However, if $f:X\to Y$ is any bijection1 whatsoever, then $f$ is an isometry between $X$ and $Y$. $\langle X,d_X\rangle$ and $\langle Y,d_Y\rangle$ are structurally identical as metric spaces: if $P$ is any property of metric spaces $-$ not just of metrizable spaces, but of metric spaces with a specific metric $-$ then either $X$ and $Y$ both have $P$, or neither of them has $P$. There is no structural property of metric spaces that distinguishes them.

What I just said about $X$ and $Y$ is true of isometric spaces in general: there is no structural property of metric spaces that distinguishes them. Considered as metric spaces, they are structurally identical, though they may have different underlying sets.

Isometric spaces may even have the same underlying set but different metrics. Consider the following two metrics on $\Bbb N=\{0,1,2,\dots\}$. For any $m,n\in\Bbb N$,

$$d_0(m,n)=\begin{cases} 0,&\text{if }m=n\\\\ \left|\frac1m-\frac1n\right|,&\text{if }0\ne m\ne n\ne 0\\\\ \frac1m,&\text{if }n=0<m\\\\ \frac1n,&\text{if }m=0<n\;, \end{cases}$$

and

$$d_1(m,n)=\begin{cases} 0,&\text{if }m=n\\\\ \left|\frac1m-\frac1n\right|,&\text{if }m\ne n\text{ and }m,n>1\\\\ 1-\frac1m,&\text{if }n=0\text{ and }m>1\\\\ 1-\frac1n,&\text{if }m=0\text{ and }n>1\\\\ \frac1m,&\text{if }n=1\ne m\\\\ \frac1n,&\text{if }m=1\ne n\;. \end{cases}$$

It’s a good exercise to show that $$f:\Bbb N\to\Bbb N:n\mapsto\begin{cases}n,&\text{if }n>1\\1,&\text{if }n=0\\0,&\text{if }n=1\end{cases}$$ is an isometry between $\langle\Bbb N,d_0\rangle$ and $\langle\Bbb N,d_1\rangle$. (HINT: Both spaces are isometric to the space $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual metric.) Yet these are clearly not the same space: metric $d_0$ makes $0$ a limit point of the other points, but metric $d_1$ makes $0$ an isolated point.

I don’t know of any general method for finding an isometry between isometric spaces; if you can recognize two spaces as being isometric, you probably already have a good idea of what an isometry between them must look like.

1 If you want a specific bijection, $$f(n)=\begin{cases}0,&\text{if }n=0\\\\\frac{n}2,&\text{if }n>0\text{ and }n\text{ is even}\\\\-\frac{n+1}2,&\text{if }n\text{ is odd}\end{cases}$$ does the job.

-
Sir, by structural properties do you mean completeness, boundedness etc? – srijan May 28 '12 at 6:05

1
@srijan: Anything that has to do with the topological or metric structure of the space and not with superficial charactersitiscs like the specific names attached to the points. Completeness and boundedness are indeed structural properties of metric spaces, though not of metrizable spaces. – Brian M. Scott May 28 '12 at 6:07
All homeomorphisms need not be isometries because some of the topological properties may not be shared by two isometric spaces, Am i right sir? – srijan May 28 '12 at 6:15

1
@srijan: No, it’s because some of the metric properties may not be shared between two homeomorphic spaces. For instance, $\Bbb R$ and $(0,1)$ with the usual metrics are homeomorphic, but $\Bbb R$ is a complete metric space, while $(0,1)$ isn’t: they don’t share the metric property of completeness. – Brian M. Scott May 28 '12 at 6:18

1
@srijan: You’re very welcome! – Brian M. Scott May 28 '12 at 6:22
show 1 more comment
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9392446875572205, "perplexity_flag": "head"}