8
$\begingroup$

This is inspired by this question.

Using invariance of domain or some such theorem, it is easy to prove there is no continuous bijection $f:\mathbb{R}^n\to [0,1]^m$ for any $n,m\in\mathbb{N}$. Otherwise, composing it with the inclusion in $\mathbb{R}^m$, we would get that $[0,1]^m$ must be open in $\mathbb{R}^m$.

However, I imagine there must be a much simpler proof using simpler properties of the cube and the Euclidean space. For instance, it's very easy to prove this for $m=1$ and any $n$, as in the question I linked to.

Any ideas?

EDIT. Two answers have been so far suggested, but Kevin's is incomplete, and I believe the other is incorrect.

$\endgroup$
6
  • $\begingroup$ Hi Cronus, I have a thought but I've not seriously check its validity. Suppose $\exists$ such $f$, then $\exists x_0=(a_1,a_2,...a_n)$ s.t. $f(x_0)=(1,1,1,....,1)$, since $f$ is continuous, $\exists U$ open neighborhood of $x_0$ s.t. $f(U)=(k,1]^m$, so for any $y\in U, y\neq x_0$, then there must exist repetition. The argument for this part is what I need to consider more, but before that, we can roughly discuss the idea. $\endgroup$
    – Kevin.S
    Commented Jul 23, 2020 at 9:34
  • 1
    $\begingroup$ @Kevin.S Nice idea, but I think it's nontrivial to assure the image of a neighrbourhood of $x_0$ contains a neighbourhood of $(1,...,1)$. $\endgroup$
    – Cronus
    Commented Jul 23, 2020 at 9:36
  • $\begingroup$ Oh! I see that problem. thanks to point it out. $\endgroup$
    – Kevin.S
    Commented Jul 23, 2020 at 14:11
  • $\begingroup$ Why can't you use compactness? $\Bbb{R}^n$ is non-compact in the standard topology but $[0, 1]^m$ is. A continuous bijection (homeomorphism) must preserve compactness. QED. $\endgroup$ Commented Jul 24, 2020 at 0:01
  • 2
    $\begingroup$ @RiversMcForge Usually the argument is that the image of a compact set under a continuous map is still compact. But here it is not assumed that the inverse of $f$ is continuous... It's just a continuous bijection, not a homeomorphism. $\endgroup$
    – WhatsUp
    Commented Jul 24, 2020 at 0:33

1 Answer 1

1
$\begingroup$

I improved the my previous idea, but it's too long to write a comment, so I decided to post it here. If anything goes wrong, I will delete my answer or edit it. Actually, I'm suspicious that I missed something because this question seems to be more complicated....

Assume $(\Bbb{R}^n,d_1)$ and $([0,1]^m,d_2)$ be the two metric spaces.

Suppose $\exists$ such $f$ and Let $x_0=(a_1,a_2,...,a_n)\in\Bbb{R}^n$ s.t. $f(x_0)=(1,1,...,1)$. Then, since both spaces are metric spaces, we see that $d_1(x_0,y)<\epsilon\implies d_2(f(x_0),f(y))<\delta$ by continuity. So, in general, we can take two sets $U,V$ containing $x_0$, where $U=\prod_{i=1}^{n-1}(a_i-\epsilon,a_i+\epsilon)\times(a_n-\epsilon,a_n]$ and $V=\prod_{i=1}^{n-1}(a_i-\epsilon,a_i+\epsilon)\times[a_n,a_n+\epsilon)$. Their images $f(U),f(V)\subset(1-\delta,1]^m$.

Case 1: If $f(U)\cap f(V)=\varnothing$

Suppose it's possible, then each point in $f(U\cap V)=f(\prod_{i=1}^{n-1}(a_i-\epsilon,a_i+\epsilon)\times a_n)$ would have two distinct images when it's in $U$ and in $V$, which contradicts the assumption that $f$ is bijective.

Case 2: If $f(U)\cap f(V)\cap ((1-\delta,1]^m\setminus f(U\cap V))\neq\varnothing$

Then there must exist repetitions that undermines the assumption about bijectivity.

Case 3: If $f(U)\cap f(V)=f(U\cap V)$

Then we let $$U=\prod_{j=1}^{n-2}(a_j-\epsilon,a_j+\epsilon)\times (a_{n-1}-\epsilon,a_{n-1}]\times a_n$$ and $$V=\prod_{j=1}^{n-2}(a_j-\epsilon,a_j+\epsilon)\times [a_{n-1},a_{n-1}+\epsilon)\times a_n.$$ By doing this, we restricts the image to an $(m-1)$ dimensional cube containing $(1,1,...,1)$, so we face exactly the same three situations again, and a similar argument push this to the $(m-2)$ dimension, where $U$ becomes $$\prod_{j=1}^{n-3}(a_j-\epsilon,a_j+\epsilon)\times (a_{n-2}-\epsilon,a_{n-2}]\times\prod_{j=n-1}^na_j$$ and $V$ becomes $$\prod_{j=1}^{n-3}(a_j-\epsilon,a_j+\epsilon)\times [a_{n-2},a_{n-2}+\epsilon)\times\prod_{j=n-1}^na_j.$$

We repeat that process with new $U,V$ on and on until the images get to dimension one, which reduces to the situation of $\Bbb{R}^n\to l\approx[0,1]$ (restricting the range to a line segment). Their images, (when restricted to $l$), should be $(a,1]$ and $(b,1]$ (the notation is a little bit weird... but should be half open segments) respectively, where $1-\delta<a,b<1$. then the intersection must be nontrivial and contain extra points.

I missed the space-filling curve, as Cronus pointed out in his comment. The main obstruction of this method is that I've not found any way to argue the fractal curves. I'm still thinking, but maybe someone could help me out.

I thought about the reason why we need to reduce the dimensions: if $m>1$, then we always get a plane or something that $U$ and $V$ may not be able to intersect as desired because the image has various directions to go, but in $m=1$, there is only one direction, so the image of $U$ and $V$ will certainly intersect each other that contains extra points.

I know that this part is hard to understand. It's also hard to describe clearly, so I think we may need further discussion on that......


I'm not really sure if this is valid, but it'll never get verified and edited unless I post it. If someone found flawness in this answer, please kindly point out and I'll reconsider or delete my post.


Edit:

Actually, if both spaces were equipped with the order topology built on the usual dictionary order, then this would be easy.

Assume $f$ is conts bijective, then $\exists x,y\in\Bbb{R}^n$ s.t. $f(x)=(0,...,0)$, $f(y)=(1,...,1)$. Because $\Bbb{R}^n$ is connected and $[0,1]^m$ is assumed to be an ordered set in the order topology, we can apply the (generalized version) intermediate value theorem: Since in the order topology, the interval $(f(x),f(y))$ must go through all points in the cube, which corresponds to the image of the interval $(x,y)$. Now, the point in $\Bbb{R}^n\setminus (x,y)$ needs to have image but it will undermines the bijectivity, which gives us a contradiction.

$\endgroup$
6
  • $\begingroup$ Thanks for taking the time to write this. It's a bit long so it might take me a while before I have the time to read it, but I will. :-) $\endgroup$
    – Cronus
    Commented Jul 23, 2020 at 14:19
  • $\begingroup$ @Cronus thank you, but be careful because I think I missed something important in this question. I'm not really confident about the case 3. : ) $\endgroup$
    – Kevin.S
    Commented Jul 23, 2020 at 14:21
  • $\begingroup$ Hmm I see. I took a look now and I think there's a mistake in this part: "We repeat that process with new U,V on and on until the images get to dimension one, which reduces to the situation of Rn→l≈[0,1] (restricting the range to a line segment)." The image of a line might not be a line; see here: en.wikipedia.org/wiki/Space-filling_curve $\endgroup$
    – Cronus
    Commented Jul 23, 2020 at 15:28
  • $\begingroup$ But your idea might still work $\endgroup$
    – Cronus
    Commented Jul 23, 2020 at 15:31
  • $\begingroup$ @Cronus I see what you meant. The so-called "line" is not necessarily a line. The main idea that I was trying to show is that because $(1,...,1)$ must be the end point while $x_0$ is an interior point, now if we are able to restrict the range to a one dimensional object, then we can say that there must exist repetition that undermines bijectivity. But now, I will need a ligetimate way of deduction. sorry for making such a mistake. $\endgroup$
    – Kevin.S
    Commented Jul 23, 2020 at 22:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .