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For a metric space $(X,d)$, let $\DeclareMathOperator{\Iso}{\operatorname{Iso}} \Iso(X, d)$ be a set of all isometries on $X$. (Function $f:X \rightarrow X$ is isometry if $d(f(x), f(y)) = d(x, y)$, for all $x,y \in X$).

Is there an example of compact metric space $(X,d)$ such that $X$ is uncountable set and $\operatorname{card}(\Iso(X,d)) = \aleph_0$?

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  • $\begingroup$ I dont think I understand, unit circle has uncountably many isometries. $\endgroup$
    – Goki
    Dec 8, 2023 at 19:57
  • $\begingroup$ the example wasn't compact $\endgroup$
    – C Squared
    Dec 8, 2023 at 19:57
  • $\begingroup$ Do you need countably infinite isometries or just contable? $\endgroup$
    – Surge
    Dec 8, 2023 at 20:38
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    $\begingroup$ Example where there is only finitely many isometries is not very hard to find. Segment [0,1] has only two. Im interested in countably infinite example. $\endgroup$
    – Goki
    Dec 8, 2023 at 20:47
  • $\begingroup$ I think the Koch snowflake curve is likely to be an example. $\endgroup$
    – Rob Arthan
    Dec 8, 2023 at 20:56

2 Answers 2

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No, there is no such example. In fact we do not even need to assume $X$ is uncountable.


Update.

$\newcommand{\isom}{\operatorname{Isom}}$ A quicker way to prove this than the original proof below is to observe that by Arzelà–Ascoli $\isom(X)$ is compact. If it is infinite, then it is not discrete, and then since the $\sup$ metric is invariant under postcomposition by isometries (i.e, $d(f\circ g_1,f\circ g_2)=d(g_1,g_2)$), $\isom(X)$ has no isolated points* (see clarification below).

Since it is also compact, hence complete, $\isom(X)$ is perfect, and therefore uncountable (and in fact has cardinality at least $2^{\aleph_0}$).


Original proof.

To see this, first note that if $X$ is compact, the isometry group $\isom(X)$ is a compact subset of the space $\mathcal C(X,X)$ under the uniform metric $d(f,g)=\sup_{x\in X} d(f(x),g(x))$, by Arzelà–Ascoli.

It follows that if $\isom(X)$ is infinite, it cannot be discrete, and thus has members arbitrarily close to the identity $I$.

We may therefore pick a sequence $f_n\in \isom(X)$ with $f_n\neq I$ and $d(f_{n+1},I)\leq \frac{1}{4}d(f_n,I)$ for each $n$.

Now for a finite subset $S=\{n_1,\dots, n_k\}\subseteq \mathbb N$, with $n_1< \dots <n_k$, define $f_S=f_{n_k}\circ\dots\circ f_{n_1}$, and then for arbitrary subsets $S\subseteq \mathbb N$, let $f_S= \lim_{n\to\infty} f_{S\cap\{1,\dots,n\}}$.

It is easy to see that each $f_S$ is an isometry. Moreover, for each $k$ we have $$d(f_{S\cap\{1,\dots,k\}},f_S)\leq \sum_{n=k+1}^{\infty} d(f_n,I)\leq \sum_{n=1}^\infty \frac{1}{4^n}d(f_k,I)=\frac{1}{3} d(f_k,I).$$

Then if $S\neq T\subseteq\mathbb N$, and $k$ is the first integer for which $S$ and $T$ differ, then

\begin{align*} d(f_S,f_T) &\geq d(f_{S\cap\{1,\dots,k\}},f_{T\cap\{1,\dots,k\}})- d(f_{S\cap\{1,\dots,k\}},f_S) - d(f_{T\cap\{1,\dots,k\}},f_T)\\ &\geq d(f_{S\cap\{1,\dots,k\}},f_{T\cap\{1,\dots,k\}})- \frac{2}{3} d(f_k,I)\\ &=d(f_k,I)-\frac{2}{3}d(f_k,I)\\ &=\frac{1}{3}d(f_k,I)>0. \end{align*}

It follows that every distinct subset $S\subseteq \mathbb N$ gives rise to a distinct isometry $f_S$, so the isometry group is uncountable.


Remark.

Throughout both answers above, I tacitly assumed we were talking about surjective isometries, whereas in the original question isometries were not defined to be a priori surjective. However, this is not a problem, since when $X$ is compact isometries are always surjective.


Additional clarification.

Since the uniform metric is preserved by post-composition of isometries, for each $f\in \isom(X)$ the map $g\mapsto f\circ g$ is an isometry of $\isom(X)$, and a fortiori is a homeomorphism, which therefore takes isolated points to isolated points. Thus if there is a single isolated isometry $h\in \isom(X)$, then for each $f\in\isom(X)$ the map $g\mapsto f\circ h^{-1}\circ g$ is a homeomorphism taking $h$ to $f$, so $f$ is isolated.

We conclude that $\isom(X)$ either has no isolated points, or is discrete. Since the latter is impossible for an infinite compact set, we have no isolated points.

Note also that this isn’t really about isometries in particular, and is true for any topological group, but I have avoided going into that discussion in case you have not yet been exposed topological groups.

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  • $\begingroup$ Nice, I'm only not able to see why from invariance under postcomposition follows Iso(X) has no isolated points. Can you please explain in a bit more detail? $\endgroup$
    – Goki
    Dec 9, 2023 at 9:23
  • $\begingroup$ @Goki added some clarification to the end on that issue, hope that helps. $\endgroup$
    – M W
    Dec 9, 2023 at 10:32
  • $\begingroup$ Thank you, I understand it now. $\endgroup$
    – Goki
    Dec 9, 2023 at 10:58
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Here is one example. Let $X \subset \mathbb{R}^2, X = \{(x,y): (x,y)\in \mathbb{R}^2, xy = 0, |x| \leq 1, |y| \leq 1\}$. This set is also compact. Let $d$ be the Euclidean metric. Obviously, $X$ is uncountable, but there are only eight isometries, corresponding to the rotations by multiples of $\pi/2$ angles around the origin, perhaps combined with a reflection around the origin. Sets like $Y = \{(x,y) \in \mathbb{R}^2: |x| \leq 1, |y| \leq 1\}$ also work.

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    $\begingroup$ 8 is strictly less than $\aleph_0$ $\endgroup$ Dec 8, 2023 at 20:36
  • $\begingroup$ Aren't all finite sets countable by definition? $\endgroup$
    – Surge
    Dec 8, 2023 at 20:37
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    $\begingroup$ Some people use countable to mean countable infinite. You should read the question and not just the title. "Card(Iso(X)) $= \aleph_0$..." $\endgroup$ Dec 8, 2023 at 20:38
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    $\begingroup$ If the OP is happy with a finite number of isometries then why not just take the closed unit interval, which admits just $2$ isometries. $\endgroup$
    – Rob Arthan
    Dec 8, 2023 at 20:48
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    $\begingroup$ The set $Z$ you have introduced in your edit isn't closed, so how can it be compact? (I assume you mean "union" rather than "interesection" as the intersection of those sets comprises a single point.) $\endgroup$
    – Rob Arthan
    Dec 8, 2023 at 20:49

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