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I have a problem but I don't know if there is a solution or a counter-example.

Problem: Let $M$ be a non trivial compact connected metric space and let $f:M\to M$ be a homeomorphism. Show that there are $a\neq b$ such that $d(f(a),f(b))=d(a,b)$.

Thanks.

Rudy the Reindeer
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Sigur
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    Thiago: I've deleted my attempt in which I tried to use that $g \colon (a,b)\mapsto d(f(a),f(b))-d(a,b)$ must attain value zero at some point. I don't know whether that reasoning can be modified to get that there is a point such that $g(x,y)\ne0$ and this point is not on the diagonal. :-( – Martin Sleziak May 18 '12 at 17:17
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    I did not read Martin's post, but the idea looks reasonable. Consider the space $M^{(2)}$ of 2-point subsets of $M$ equipped with the Hausdorff metric (this is called the symmetric product of $M$ with itself). Equivalently, it's $M\times M$ minus diagonal modulo the equivalence $(a,b)\sim (b,a)$. The function $g\colon M^{(2)}\to\mathbb R$ introduced by Martin attains values $\ge 0$ and $\le 0$ (consider the two-point sets which realize the diameter of $M$). The missing piece is the connectedness of $M^{(2)}$... I'm pretty sure that $M^{(2)}$ is connected whenever $M$ is. –  May 18 '12 at 18:54
  • Actually, I abused terminology here: the symmetric product, as defined by Borsuk-Ulam and others, includes the diagonal. If the diagonal is included, connectedness is obvious. I still think that $M^{(2)}$ should be connected even without the diagonal. –  May 18 '12 at 19:23

1 Answers1

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I'll write $|x-y|$ instead of $d(x,y)$ to lighten notation. Suppose to the contrary that $|f(x)-f(y)|\ne |x-y|$ whenever $x\ne y$. For each $x\in M$ let $E_x=\{y\in M\colon |f(x)-f(y)|>|x-y|\}$ and $C_x=\{y\in M\colon |f(x)-f(y)|<|x-y|\}$. Clearly, both $E_x$ and $C_x$ are open and $M\setminus\{x\}=E_x\cup C_x$.

A point $x$ is called a non-cut point if $M\setminus\{x\}$ is connected. Every nontrivial connected compact metric space has at least two non-cut points (see Analytic Topology by Whyburn, p.54). Let $N$ be the set of all non-cut points of $M$, and let $d$ be its diameter, $d>0$. For every $x\in N$ either $E_x$ or $C_x$ is equal to $M\setminus\{x\}$.

Case 1. $E_x=M\setminus \{x\}$. In particular, for every other non-cut point $y$ we have $|f(x)-f(y)|>|x-y|$. Hence $E_y$ is nonempty, which implies $E_y=M\setminus\{y\}$ because $M\setminus\{y\}$ is connected. In other words, $f$ increases all distances between non-cut points.

Case 2. $C_x=M\setminus \{x\}$. Same reasoning as above shows that $f$ decreases all distances between non-cut points. Since $f$ can be replaced with $f^{-1}$, in the sequel we assume that Case 1 holds.

Choose two sequences $(x_n)$ and $(y_n)$ in $N$ such that $|x_n-y_n|\to d$. By passing to subsequences we may assume that both converge: $x_n\to x$ and $y_n\to y$. Clearly, $|x-y|=d$. Since $f$ is a homeomorphism, it maps $N$ onto itself. Thus $d\ge |f(x_n)-f(y_n)|>|x_n-y_n|$ for all $n$. Passing to the limit we obtain $|f(x)-f(y)|=\lim_{n\to\infty} |f(x_n)-f(y_n)|=d$. Thus, $|f(x)-f(y)|=|x-y|$, a contradiction.

Martin Sleziak
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    I've added added Google Books link to Whyburn's books to your post - I guess this might be useful for some users. (Of course, I am aware that the particular page will not be displayed to everyone.) I hope you don't mind. – Martin Sleziak May 21 '12 at 06:21
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    @Martin No problem. This fact is also mentioned (of course without proof) on Wikipedia, http://en.wikipedia.org/wiki/Cut-point || Actually, I'd rather have a proof via the connectedness of the set of unordered pairs (as in the comments to the original post), because it wouldn't involve the continuum theory. Unfortunately, I still can't prove the connectedness. Maybe I'll post it as a separate question since it's of independent interest. –  May 21 '12 at 13:09
  • Sorry guys, I was busy. I would like to thank Leonid for the answer. It's very interesting as the fact about cut-point as well. Martin, thanks for the book. I'll read that page. Thanks again. – Sigur May 21 '12 at 13:48
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    For those who can’t see the Google Books page, this answer to an earlier question has a link to the 1968 PNAS paper in which Whyburn published the proof that every non-degenerate compact, connected $T_1$-space has at least two non-cut points. – Brian M. Scott May 23 '12 at 08:42
  • So, the approach is no trivial. Nice paper. I guess that without that result it would be impossible to solve the problem. Thanks for the comments. – Sigur May 23 '12 at 16:05
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    @Thiago There is now an alternative proof, which does not use the existence of non-cut points. Namely, $C_2(X)$ is connected for any connected compact Hausdorff space $X$: see http://math.stackexchange.com/a/152345/31373 and the reference therein (and the reference therein). –  Jun 02 '12 at 18:10
  • Another update: the same thread now has a self-contained proof that $C_2(X)$ is connected for any connected topological space $X$. –  Jun 04 '12 at 23:17
  • @MartinSleziak: ¿why x and y are non-cut points of M? –  Jul 08 '13 at 15:30
  • @NéstorRaúlPachón We do not need that $x,y\in N$. (And I do not think this is true in general.) It suffices to show that $x,y\in M$ and $d(x,y)=d(f(x),f(y))$. – Martin Sleziak Jul 09 '13 at 06:18