36
$\begingroup$

I need a hint. The problem is: is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$

I'm pretty sure that there aren't any, but so far I couldn't find the proof.

My best idea so far is to consider $f' = f|_{\mathbb{R}-\{*\}}: \mathbb{R} - \{*\} \to \mathbb{R}^2 - \{f(*)\}$, and then examine the de Rham cohomologies: $$H^1_{dR}(\mathbb{R}^2 - \{f(*)\}) = \mathbb{R} \ \xrightarrow{H^1_{dR}(f')} \ 0 = H^1_{dR}(\mathbb{R} - \{*\}),$$ but so far I failed to derive a contradiction here. Am I on the right path? Is it possible to complete the proof in this way e.g. by proving that $H^1_{dR}(f')$ must be a mono? Or is there another approach that I missed?

$\endgroup$
7
  • 1
    $\begingroup$ What about a Hilbert curve? $\endgroup$
    – jimjim
    Jun 25, 2011 at 6:58
  • 7
    $\begingroup$ @Arjang: Those are not injective. $\endgroup$
    – t.b.
    Jun 25, 2011 at 6:59
  • 1
    $\begingroup$ @Theo : Thanks, I always imagined the limiting case of hilbert curve to be a mapping from $\mathbb R to \mathbb R^2$ , I feel disillusioned that is not the case. is it a question worth asking? $\endgroup$
    – jimjim
    Jun 25, 2011 at 7:08
  • 1
    $\begingroup$ @Arjang: It's your call. But why don't you try looking into it yourself? Usually they are from the unit interval onto to the unit square and if such a map were injective, it would be a homeomorphism, which it can't be for reasons of connection (removing an interior point of the interval disconnects it, whereas ...) $\endgroup$
    – t.b.
    Jun 25, 2011 at 7:14
  • 2
    $\begingroup$ arxiv.org/pdf/1003.1467.pdf $\endgroup$
    – user27182
    Mar 26, 2013 at 1:00

3 Answers 3

Reset to default
36
$\begingroup$

Suppose $f(x)$ were such a function. Note that each $A_n = f([-n,n])$ is a closed (actually compact) set, with $\cup A_n = {\mathbb R}^n$. By the Baire category theorem, there is one such $A_n$ that contains a closed ball $B$. Since $[-n,n]$ is compact, the image of any relatively closed subset of $[-n,n]$ is compact and thus closed. Hence $f^{-1}$ is continuous when restricted to $A_n$, and thus when restricted to $B$. So in particular $f^{-1}(B)$ is a connected subset of ${\mathbb R}$. Since all connected subsets of ${\mathbb R}$ are intervals, $f^{-1}(B)$ is a closed interval $I$.

Let $x$ be any point in the interior of $B$ such that $f^{-1}(x)$ is not an endpoint of $I$. Then $B - \{x\}$ is still connected, but $f^{-1}(B - \{x\})$ is the union of two disjoint intervals, which is not connected. Since $f^{-1}$ when restricted to $B - \{x\}$ is continuous, you have a contradiction.

$\endgroup$
3
  • $\begingroup$ Nice answer! But let me ask you one question: so you said that $f^{-1}(B)=I$ where $I$ is closed interval. It is closed because $f^{-1}$ is continuous and $B$ compact, right? Is it possible that $I=[a,a]$? $\endgroup$
    – RFZ
    Nov 8, 2019 at 4:56
  • $\begingroup$ Btw, where you are using that $f$ is injective and surjective? $\endgroup$
    – RFZ
    Nov 8, 2019 at 5:02
  • $\begingroup$ $I$ can't be $[a,a]$ since $f^{-1}$ is one to one; $B - \{x\}$ has more than one point, so $f^{-1}(B - \{x\})$ also has more than one point. As for where we are using injectivity and surjectivity ... even to talk about $f^{-1}$ requires one to oneness, for example. Surjectivity is used when saying $\cup_n A_n = {\mathbb R}^n$. $\endgroup$
    – Zarrax
    Nov 9, 2019 at 23:45
6
$\begingroup$

Here is a hint: What simple space is $\mathbb{R}^2 - \{ f(\ast) \}$ homotopic to?

Edit: Just a small edit, to hopefully bump this up. I had read this as a homeomorphism, in which case it is easy. However we only have a continuous bijection from $\mathbb{R} \to \mathbb{R}^2$.

There may be a way to argue from the fact that $\mathbb{R} - \{ \ast \}$ is disconnected and $\mathbb{R^2} - \{ f(\ast) \}$ is connected. This will work immediately to show there is no continuous bijection from $\mathbb{R}^2 \to \mathbb{R}$ as the continuous image of a connected set is connected. I am not sure about getting something out of the other direction however (perhaps the 'simplest' is Zarrax's explanation). Hopefully the experts will have something to add!

$\endgroup$
10
  • $\begingroup$ It's a cylinder, of course. $\endgroup$
    – Alexei Averchenko
    Jun 25, 2011 at 7:29
  • 5
    $\begingroup$ @Alexei - keep homotoping! $\endgroup$
    – Juan S
    Jun 25, 2011 at 7:30
  • 3
    $\begingroup$ A circle then? :) $\endgroup$
    – Alexei Averchenko
    Jun 25, 2011 at 7:31
  • $\begingroup$ @Alexei - indeed. $\endgroup$
    – Juan S
    Jun 25, 2011 at 7:32
  • $\begingroup$ Can I have another hint? Am I to use $f'$ to construct a homotopy from $\mathbb{R} - \{*\}$ to $S^1$ or some kind of 'homotopic injection' that would translate to a mono from $H_0(\mathbb{R} - \{*\}) = \mathbb{Z}^2$ to $H_0(S^1) = \mathbb{Z}$ and thus get a contradiction, or am I to play with $f'$ as an attaching map, or is it something I still fail to see? $\endgroup$
    – Alexei Averchenko
    Jun 26, 2011 at 1:10
2
$\begingroup$

From the Wikipedia Space-filling curves page:

A non-self-intersecting continuous curve cannot fill the unit square because that will make the curve a homeomorphism from the unit interval onto the unit square (any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism). But a unit square has no cut-point, and so cannot be homeomorphic to the unit interval, in which all points except the endpoints are cut-points.

$\endgroup$

You must log in to answer this question.

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