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This question was asked to me by a mathematics undergraduate to me and I was not able to solve it. So, I am asking it here.

Prove or Disprove : There exists a continuous bijection from $\mathbb{ R}^2$ to $\mathbb{R} $ .

I have no idea on how this problem can be tackled. It seems it has something to do with set theory but I only know elementary set theory( bijection from naturals) and I am unable to solve it.

YuiTo Cheng
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4 Answers4

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The standard argument that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}$ works just as well to show that there is no continuous bijection $\mathbb{R}^2\to \mathbb{R}$.

Suppose such a continuous bijection $f$ exists. If we remove a point $p$ from $\mathbb{R}$, it becomes disconnected. The preimages of the disjoint open sets $(-\infty,p)$ and $(p,\infty)$ will be disjoint open subsets of $\mathbb{R}^2$ whose union is $\mathbb{R}^2\setminus \{f^{-1}(p)\}$. But this implies that $\mathbb{R}^2$ minus a point is disconnected, which is a contradiction, since $\mathbb{R}^2$ minus a point is path connected.

It is also true that there is no continuous bijection $\mathbb{R}\to \mathbb{R}^2$, but the proof is a bit harder - I don't know a way of proving this that doesn't use the Baire category theorem. See this answer and also this answer.

Alex Kruckman
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Stronger result: Suppose $f:\mathbb R^2\to \mathbb R.$ Define $V_x$ to be the vertical line $\{x\}\times \mathbb R.$ Assume that $f$ is continuous on each $V_x,$ (i.e., $f(x,y)$ is continuous in $y$ everywhere), and that the collection $\{f(V_x):x\in \mathbb R\}$ is pairwise disjoint.

Claim: $f$ is constant on all but countably many vertical lines.

Proof: Let $$E=\{x\in \mathbb R: f \text{ is nonconstant on }V_x\}.$$ Then $f(V_x)$ is an interval of positive length for $x\in E.$ These intervals are pairwise disoint. Thus there can be no more than countably many of them (each such interval contains a rational, there are only countably many rationals). This is the desired result.

zhw.
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  • How would you prove that number of pairwise disjoint intervals are uncountable? –  Jun 28 '20 at 15:05
  • There are uncountably many vertical lines in the plane. – zhw. Jun 28 '20 at 15:14
  • I edited my answer; the result is stronger now. – zhw. Jun 28 '20 at 18:55
  • I like this more than my own A for its simplicity and lack of necessary background knowledge. A variant would be to consider ${f(C_r): r\in \Bbb R^+}$ where $C_r={(x,y)\in \Bbb R^2: x^2+y^2=r^2}.$ – DanielWainfleet Jun 30 '20 at 00:34
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The answer is NO!

If possible let $f :\Bbb R^2 \to \Bbb R$ be your continuous bijection.

Take any two points say $p,q \in \Bbb R^2$. Join them by a straight line segment say $l_{p,q}$ and then look at $f(l_{p,q})$. You can parametrize your $l_{p,q}$ from some $[a,b] \subset \Bbb R$ by some $\gamma$ . Then look at $f \circ \gamma :[a,b] \to \Bbb R$, it's a contiuous function from $\Bbb R \to \Bbb R$ and hence by the intermediate value theorem you get that every point in the connected segment $[f(p),f(q)]$ has a pre-image on $l_{p,q}$.

But $l_{p,q}$ isn't anything special! Same argument holds for any non self-intersecting path joining $p,q $ in $\Bbb R^2$. But then you realize that there are infinitely many non self-intersecting paths in $\Bbb R^2$ which are each disjoint to all the other ones except at the end-points $p,q$. Hence every interior point in $[f(p),f(q)]$ has infinitely many pre-images!

So, I am just trying to show that any continuous function $ \Bbb R^2 \to \Bbb R$ is so far from being injective! Uncountably many disjoint paths get mapped to every interval and every point has uncountably many pre-images!

Brozovic
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  • Nice argument. I don't see the point of the parametrization by $[a,b]$, though. It's true by definition of the image $f(I_{p,q})$ that every point of $f(I_{p,q})$ has a pre-image on $I_{p,q}$. – Alex Kruckman Jun 27 '20 at 17:36
  • I think instead the crucial thing you're using is that $f(I_{p,q})$ contains the interval $[f(p),f(q)] \subseteq \mathbb{R}$. And this is because the continuous image of a connected compact set is connected and compact, hence is a closed interval in $\mathbb{R}$. – Alex Kruckman Jun 27 '20 at 17:37
  • @AlexKruckman Since one only has IVT on functions $\Bbb R \to \Bbb R$ and not from $\Bbb R^2 \to \Bbb R$ – Brozovic Jun 27 '20 at 17:37
  • Oh, I see - again, it's trivial that every point in $f(I_{p,q})$ has a preimage on $I_{p,q}$. What you're really doing is using using IVT to prove that every point in $[f(p),f(q)]$ has a preimage on $I_{p,q}$, i.e. that $[f(p),f(q)]\subseteq f(I_{p,q})$. It's nice that one can use IVT to prove this in this case, rather than using the more complicated general topology argument in my second comment above. – Alex Kruckman Jun 27 '20 at 17:39
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    @AlexKruckman Exactly. I am just trying to show that any continuous function $ \Bbb R^2 \to \Bbb R$ is so far from being injective! Uncountably many disjoint paths get mapped to every interval and every point has uncountably many pre-images! – Brozovic Jun 27 '20 at 17:43
  • I edited to replace $f(I_{p,q})$ by $[f(p),f(q)]$ and make the appeal to IVT clear. I hope you don't mind! – Alex Kruckman Jun 27 '20 at 17:47
  • @AlexKruckman It's fine! In fact thanks for the edit, you added clarity in the answer for everyone. – Brozovic Jun 27 '20 at 17:51
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Suppose $f:\Bbb R^2\to \Bbb R$ is a continuous injection. Let $S=[0,1]^2$ and $T=[1/3,2/3]^2.$ Each of $S,T$ is compact and connected so their images $f[S], f[T]$ are compact and connected.

$(\bullet)\,$ So $f[S]$ and $f[T]$ are closed bounded real intervals.

Now $f|_S: S\to f[S]$ is a continuous bijection from the compact Hausdorff space $S$ to the compact Hausdorff space $f[S]$ so $f|_S:S\to f[S]$ is a homeomorphism.

Therefore $f$ maps the boundary of $T$ in the space $S$ to the boundary of $f[T]$ in the space $f[S].$

But the boundary of $f[T]$ in the space $f[S]$ contains just 2 points by $(\bullet)$ and the boundary of $T$ in the space $S$ is infinite, and $f$ is 1-to-1, which is absurd.

DanielWainfleet
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