Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose X is a complete metric space, and $f:X↦X$ a continuous surjective function. Let D be a dense set. Suppose $f:D↦D$ is injective and $f^{-1}(D)=D$.

Is $f$ injective ?

Is there a family of metric spaces where you can conclude $f$ is injective?

share|improve this question
3  
nice question... why should we close it? –  Pietro Majer Oct 31 '11 at 8:16
    
@Joel: is it obvious that one can `symmetrize' the examples in the question you link to so as to apply to the present question? –  Yemon Choi Oct 31 '11 at 8:27
    
Related: mathoverflow.net/questions/73719/… –  Joel David Hamkins Oct 31 '11 at 8:28
1  
Yemon, no, I guess not. This question is more interesting! –  Joel David Hamkins Oct 31 '11 at 8:47
    
Joel, I understand from the text that $f$ splits into $f_{|D}:D\to D $ which is bijective and $f_{|X\setminus D}:X\setminus D\to X\setminus D $ which is surjective. –  Pietro Majer Oct 31 '11 at 11:49
add comment

6 Answers

up vote 8 down vote accepted

Sorry but I could not resist:

alt text

The map is defined on $[0,1]\times[0,1]$, and can be written as $$ f(x,y) = \begin{cases} (x, (2-x)y) & \text{ if }y\leq 1/2, \\ (x, xy+1-x) & \text{ if }y> 1/2. \end{cases} $$

Just take $D$ to be the set of $(x,y)$ with rational coordinates in $(0,1)\times(0,1)$, then $f$ is bijective on $D$ (because it can be easily inverted), but it is clearly not injective on $[0,1]\times[0,1]$.

share|improve this answer
add comment

Here is another example which is easy to visualize. Let $X=[-1,1]^2$, $I$ be the segment between $(-1,0)$ and $(1,0)$ and $I_0$ its subsegment between $(-1/2,0)$ and $(1/2,0)$. There is a homeomorphism $f$ between $X\setminus I_0$ and $X\setminus \{0\}$ which sends $I\setminus I_0$ to $I\setminus\{0\}$ and extends continuously by setting $f(I_0)=0$. (Think of contracting the segment $I_0$ to the center while deforming the rest of the square continuously.) This map can be given by an explicit formula which is piecewise linear on each horizontal segment. Now define $D=X\setminus I$ and we are done.

share|improve this answer
add comment

The answer to the second question is "yes, for discrete spaces where metric is $d(x, y) = 1,$ for $x \neq y.$

share|improve this answer
    
This is by far the best answer to a question that I am surprised was not closed immediately. –  Bill Johnson Nov 3 '11 at 18:18
add comment

I think I can prove that if X=[0,1] (or 1-dimensional) then the claim of FelipeG is true.

Indeed, let $X=[0,1]$ and $D,f$ be as above, and proceed by contradiction. Assume there are points $a,b\in [0,1]$ such that $a<b$ and $f(a)=f(b)$; then, since $(a,b)\cap D$ has infinitely many elements, then there is some point $c\in D \cap (a,b)$ such that $f(c)=s \neq r$ (say $s>r$). On the other hand, by continuity of $f$, for every value $y\in D\cap (r,s)$ (which is not empty) there is an element $x_1\in (a,c)$ such that $ f(x_1) = y $ and an element $x_2\in (c,b)$ such that $ f(x_2)= y $. But, since $f^{-1}D=D$ both $x_1, x_2$ belong to $D$, which contradits the injectivity of $f_{|_D}$.

PS: this construction works if $f_{|_D}$ is a bijection (as one might argue from the title), otherwise there is a counterexample

share|improve this answer
    
I took the liberty of fixing your LaTeX -- hope you don't mind –  Yemon Choi Oct 31 '11 at 8:24
    
@Yemon Choi I was trying to fix it as well... In fact I still cannot visualize my comment correctly: may be the fact we were both working on the file simultaneously has generated some mess –  ccarminat Oct 31 '11 at 8:28
    
Let me have another go... –  Yemon Choi Oct 31 '11 at 8:28
    
[@Yemon Choi] now it works: thanks! –  ccarminat Oct 31 '11 at 8:32
    
Nice. I understand that $r:=f(a)$, right? –  Pietro Majer Oct 31 '11 at 12:41
add comment

There are counterexamples.

Let $X$ be the ordinal $\omega^2+1$, which as a topological space is the same as infinitely many convergent sequences, whose limit points converge. This space is homeomorphic to a countable closed subset of the unit interval and is therefore completely metrizable. Let $D$ be the isolated points of $X$, which is exactly the set of successor ordinals below $\omega^2$. This is dense, since the closure adds the missing limit ordinals. Let $f$ be the function that interleaves two successive sequences together into one. That, we combine the successor ordinals in the interval $[\omega\cdot 2n,\omega\cdot 2(n+1))$ to those in $[\omega\cdot n,\omega\cdot(n+1))$ by an injective function that simply interleaves the two sequences into one. This is injective on $D$ and also surjective. The function $f$ extends continuously to $X$ by mapping the limit points of the successive sequences, $\omega\cdot2n$ and $\omega\cdot2(n+1)$ both to $\omega\cdot n$, and $\omega^2\mapsto \omega^2$. Note that the extension $f$ is not injective.

A simpler version of this example, without ordinals, is to take $X$ to be the positive integers, plus a sequence converging to each of them in the interval below. Specifically, let $X$ have points $k$ and also $k-\frac 1n$ for positive integers $k$ and $n$. Thus, $X$ is a countable closed subset of $\mathbb{R}$. Let $D$ be the isolated points $k−\frac1n$, and let $f$ be the function that interleaves successive convergent sequences into one. That is, for each adjecent pair of sequences, converging to $2n$ and $2(n+1)$, we map the isolated points of the sequences bijectively to the sequence converging to $n$. This is bijective on D, but extends continuously to $X$, and is not injective on $X$.

(I have removed the earlier flawed example.)

share|improve this answer
1  
I don't understand. You have $1\in D$ and $f(-1)=1$, so $-1\in f^{-1}(D)$ but $-1\not\in D$. –  Guillaume Brunerie Oct 31 '11 at 11:07
    
Oops. I go back to my first examples, where $f$ is bijective on $D$. –  Joel David Hamkins Oct 31 '11 at 12:36
add comment

Here's an example. First take a countable family $\lbrace A_n:n\in\mathbb{N}\rbrace$ of countable dense sets in $(0,1)$ and then let $f:[0,1]\to\[0,1]$ be continuous such that for each $n$ the maps $f$ is an order-isomorpism between $A_n\cap(0,1/1)$ onto $A_{2n}$ and an order-reversing isomorphism between $A_n\cap(1/2,1)$ and $A_{2n+1}$. Thus $f$ is a bijection between $D=\bigcup_{n]1}^\infty A_n$ and itself, but it is not injective on $[0,1]$. The map $f$ is readily constructed, first recursively on $D$ and then by continuous extension to all of $[0,1]$.

Addendum: to construct the restriction of $f$ to $[0,1/2]$ modify Cantor's proof of the uniqueness of $\mathbb{Q}$ to construct an order-isomorphism $g$ between $D\cap(0,1/2)$ and $E=\bigcup_{n=1}^\infty A_{2n}$ that maps $A_n\cap(0,1/2)$ onto $A_{2n}$ for each $n$. Then define $f(x)=\sup\lbrace g(d):d\in D, d < x \rbrace $. Define $f$ on $[1/2,1]$ in a similar fashion.

Another addendum: I missed/overlooked the condition that $D=f^{-1}[D]$; my map is a bijection from $D$ to itself and that's it. The near-homeomorphisms on the square settle it nicely. In the spirit of those constructions one can get a one-dimensional example on the $\sin\frac1x$-curve: in every arc between points with coordinates $((n+\frac12)\pi)^{-1}$ and $((n+\frac32)\pi)^{-1}$ shrink the interval of points with $y$-coordinates between $-1/2$ and $1/2$ to the interval $[-1/n,1/n]$, everything in a bijective way. Then on the limit segment $\lbrace0\rbrace\times[-1,1]$ the interval $[-1/2,1/2]$ is collapsed to a point. Here $D$ is the graph of $\sin\frac1x$ of course.

share|improve this answer
    
I don't understand how you can extend your function to $[0,1]$, you will need uniform continuity. –  Guillaume Brunerie Oct 31 '11 at 0:00
    
I think your counterexample contains a bug (probably the problem is in the continuous extension). In fact I think I can prove that if X=[0,1] (or 1-dimensional) then the claim of FelipeG is true. –  ccarminat Oct 31 '11 at 8:03
    
---I think I can prove that if X=[0,1] (or 1-dimensional) then the claim of FelipeG is true--- That's unlikely. Take $f:x\mapsto 2x^2-1$ on $[-1,1]$ and take some dense set of transcendental algebraicly independent numbers. Now, for each number take its full forward orbit and some its backward orbit (say, choosing the preimage from $[0,1]$ at each step). Clearly, all numbers you get will be different, so $f$ is one to one on the resulting dense set $D$. A similar construction should work in all but very degenerate metric spaces. –  fedja Oct 31 '11 at 13:20
    
Fedja, if you only take some of the pre-image, then you won't have $f^{-1}(D)=D$. –  Joel David Hamkins Oct 31 '11 at 13:26
1  
Ah, right you are! I somehow missed that part of the condition entirely. Shame on me! –  fedja Oct 31 '11 at 17:59
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.