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

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.

• Any continuous and bijective z=f(x,y) would do – Divide1918 Jun 27 '20 at 17:22
• Perhaps this could prove to be helpful: math.stackexchange.com/questions/183361/… – Omar S Jun 27 '20 at 17:22
• The answer is "no". It probably will take some knowledge of point-set topology to answer this. – GEdgar Jun 27 '20 at 17:23

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.

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.

• How would you prove that number of pairwise disjoint intervals are uncountable? – No -One 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 0:34

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!

• 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
• @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

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.

• It is also true that there is no continuous injection $f:\Bbb R^n\to \Bbb R^m$ when $2\le m<n\in \Bbb N$ but not so easily. – DanielWainfleet Jul 5 '20 at 1:44