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Let $(X,d)$ be a compact metric space. Let $f: X \to X$ be such that $d(f(x),f(y)) = d(x,y)$ for all $x,y \in X$. Show that $f $ is onto (surjective).

If $f$ is not onto then there exist a $p \in X$ such that there does not exist any $y \in X$ such that $f(y) =p$. Then there exist $x \in X$ such that $d(p,f(x)) = d(p,x)$.

I am finding difficulty to do the proof please help!!

Is the result true if $X$ is not compact??

User8976
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  • What's a compact metric space? (I can guess it'd be the subspace topology applied to a compact subspace, which is metricisable - but I'd like to hear this) – Alec Teal Feb 13 '15 at 16:23
  • @AlecTeal Sequential compactness can be formulated without reference to an ambient space. Completeness and total boundedness can also be defined without reference to an ambient space. – Ian Feb 13 '15 at 16:48
  • @Ian is sequential something very different to normal compactness, because "normal" compactness requires an ambient space. – Alec Teal Feb 13 '15 at 16:53
  • @AlecTeal It is true that sequential and compactness are rather different notions, although in a metric space they turn out to be equivalent. Also, I think one can indeed formulate the notion of a "compact topological space" without regard to an ambient space. Specifically, a topological space $X$ is compact if every $X$-open cover of $X$ has a finite subcover. – Ian Feb 13 '15 at 16:58
  • @AlecTeal (Cont.) Moreover, a subset $A$ of an ambient space $X$ is compact (as a subset of $X$, meaning all $X$-open covers of $A$ have a finite subcover) if and only if $A$ with the subspace topology from $X$ is compact (as a subset of itself, meaning all $A$-open covers of $A$ have a finite subcover). These are trivially equivalent, because ${ U_\alpha }$ is a cover if and only if ${ U_\alpha \cap A }$ is a cover, and $ U_\alpha \cap A $ will be open in $A$ if $U_\alpha$ was open in $X$. Are you sure this doesn't work? – Ian Feb 13 '15 at 17:03
  • @Ian I didn't say something didn't work. I just wasn't sure what sequential compactness was, I know you need an ambient space to talk about compactness and you said that sequential compactness does not require an ambient space, so I asked if they were something very different. – Alec Teal Feb 13 '15 at 17:23
  • @AlecTeal Oh, I see now, you put a period instead of a question mark, so I assumed you meant "sequential [compactness] is something very different from normal compactness." Anyway yes, topological compactness doesn't require an ambient space, and the notion of a compact space (where the covers come from the space itself) and a compact subset (where the covers come from the ambient space) coincide. My apologies for the confusion. – Ian Feb 13 '15 at 17:24
  • @Ian I made a statement "Compactness requires an ambient space" - which you know (because you defined it) - what is sequential? – Alec Teal Feb 13 '15 at 17:26
  • @AlecTeal Compactness doesn't require an ambient space, even in the general topological setting; this is what I've been saying for the last few comments. (I made a mistake initially in thinking that it was similar to the notions of closure, boundary, etc.) Sequential compactness (which is what is useful in the OP's problem) means that every sequence has a convergent subsequence. – Ian Feb 13 '15 at 17:28
  • @AlecTeal For example, a closed interval $[a,b]$ is compact in itself because any cover by sets which are open in $[a,b]$ must contain intervals of the form $[a,a+\delta_1)$ and $(b-\delta_2,b]$ for some (perhaps small) $\delta_1,\delta_2>0$. These sets are open in $[a,b]$ itself despite not being open in $\mathbb{R}$. – Ian Feb 13 '15 at 17:30
  • @Ian http://www.maths.kisogo.com/index.php?title=Compactness one sec - doing the proof – Alec Teal Feb 13 '15 at 17:38
  • @AlecTeal A better way of thinking about it: suppose $X$ is a topological space and $A$ is a subset of $X$. Then $A$ is compact in $X$ if every $X$-open cover of $A$ has a finite subcover. $X$ itself is compact if it is compact (in the previous sense) in itself. The latter notion is what is meant by a "compact metric space". These are ultimately the same, because if $A$ is compact in $X$ then $A$ with the subspace topology from $X$ is compact in itself. – Ian Feb 13 '15 at 19:10
  • The last part of this solutions, solves your problem : http://math.stackexchange.com/questions/886944/isomorphism-isometries-between-finite-subsets-implies-isomorphism-isometry-bet – Fardad Pouran Feb 14 '15 at 09:25
  • possible duplicate of Isometries of $\mathbb{R}^n$ – Najib Idrissi Mar 14 '15 at 15:13

4 Answers4

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Consider a point $p\in X$, and define the sequence $(x_n)_n$ inductively by setting $x_0=p$ and $x_{n+1}=f(x_n)$. Since $X$ is compact there exists a convergent sub-sequence $(x_{n_k})_k$. In particular we have $$\lim_{k\to\infty}d(x_{n_{k+1}},x_{n_k})=0$$ That is $$\lim_{k\to\infty}d(f^{n_{k+1}}(p),f^{n_{k}}(p))=0$$ Or, using the assumption $$\lim_{k\to\infty}d(f^{n_{k+1}-n_k}(p),p)=0$$ Since $n_{k+1}-n_k\ge1$ the above result is equivalent to $$\lim_{k\to\infty}d(f(y_k),p)=0\tag{1}$$ where $y_k=f^{n_{k+1}-n_k-1}(p)=x_{n_{k+1}-n_k-1}$. Now we can extract from $(y_k)_k$ a convergent sub-sequence $(y_{k_m})_m$ that converges to some $q\in X$, and $(1)$ then implies, (due to the continuity of $x\mapsto d(f(x),p)$), that $$\lim_{m\to\infty}d(f(y_{k_m}),p)=d(f(q),p)=0$$ So $p=f(q)\in f(X)$. This proves that $f$ is onto.

Remark. Note that we only need that $d(f(x),f(y))\ge d(x,y)$ for every $ x,y$ in $X$.

Omran Kouba
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Suppose $f$ is not onto. Pick $x \in X \setminus f[X]$. Then let $\epsilon > 0$ be such that $\epsilon < d(x, f[X])$. $X$ can be covered by finitely many open sets of diameter $< \epsilon$, by compactness. Let $N$ be the smallest size of such a covering, and $\mathcal{U} = \{O_1,\ldots,O_N\}$ a witnessing cover. If $x \in O_i$, then $f[X]$ does not intersect $O_i$, so $f[X]$ is already covered by $\mathcal{U} \setminus \{O_i\}$, which has $N-1$ elements. Then $\{f^{-1}[O_j]: j \neq i \}$ covers $X$, consists of open sets, and the isometry property guarantees that all diameters are $< \epsilon$. This contradicts the minimality of $N$, contradiction.

Henno Brandsma
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you already got your 1st answer..for your 2nd question the result is not true if X is not compact...for an example define $f: \mathbb{N} -> \mathbb{N}$ s.t $f(n)=n+1$ is not surjective

Anubhav Mukherjee
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If $f$ is not onto, take a point $p$ not in the image of $f$. Using sequential compactness, show that there is an open ball $B_\epsilon(p)$ centered at $p$ which is not contained in $f(X)$. Next, show that it cannot be the case that $B(p,\epsilon) \cap f(X) = \emptyset$ for some $\epsilon > 0$. Suppose to the contrary that it does. Then consider the sequence $a_1 = p$, $a_{n+1} = f(a_n)$, $n\ge 1$. Show that by compactness, there are $m$ and $n$, $n > m$, such that $d(a_m,a_n) < \epsilon$. By the isometry property of $f$, $d(f^{n-m}(p), p) < \epsilon$, which contradicts $B_\epsilon(p)\cap f(X) = \emptyset$.

kobe
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