# No continuous injective functions from $\mathbb{R}^2$ to $\mathbb{R}$

Which of the following statements is true?

$$(a)$$ There are at most countably many continuous maps from $$\mathbb{R}^2$$ to $$\mathbb{R}$$

$$(b)$$ There are at most finitely many continuous surjective maps from $$\mathbb{R}^2$$ to $$\mathbb{R}$$

$$(c)$$ There are infinitely many continuous injective maps from $$\mathbb{R}^2$$ to $$\mathbb{R}$$

$$(d)$$ There are no continuous bijective maps from $$\mathbb{R}^2$$ to $$\mathbb{R}$$

My thinking :- $$(a)$$ and $$(b)$$ are false since $$f(x,y)=kx$$ where $$k\in \mathbb{R}$$ are continous and choosing $$k\neq 0$$ falsifies $$(b)$$

For $$(c)$$ and $$(d)$$, My best guess is that there is no continuous injective function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$ but I can't prove that !.

I was thinking that $$\mathbb{R}^2$$ is field isomorphic to $$\mathbb{C}$$ and so the required function is a real-valued function in complex variable but then I am clueless...

• For (c) and (d) consider what happens when you try to continuously map the points on the circumference of a circle from $\mathbb R^2$ to $\mathbb R$ and what happens between the maximum and the minimum Commented Dec 27, 2020 at 4:51
• Thanks @Henry . There are two points on the circle corresponding to which we get the maximum and minimum. Now there are two arcs joining these points but the corresponding image sets is the same interval which contradict injectivity of $f$. Have I understood your hint properly ? Commented Dec 27, 2020 at 5:11

Suppose there exists $$f:\mathbb R^2\to\mathbb R$$ injective and continuous. Since $$f$$ is continuous, its image, $$\operatorname{Im} f$$, must be a connected non-empty subset of $$\mathbb R$$, ie. an interval. Moreover, since $$f$$ is injective, said interval must contain more than one point and therefore, has an interior.

Let $$y$$ be an interior point of $$\operatorname{Im} f$$ and $$x=f^{-1}(y)$$. The set $$A=\mathbb R^2\smallsetminus\{x\}$$ is clearly connected in $$\mathbb R^2$$ but $$f(A)$$ is not (as it is an interval minus one point, $$y$$). This contradicts the continuity of $$f$$.

Your counterexamples for (a) and (b) are fine.

For (c) and (d) one can make a cardinality argument. suppose that $$f:\Bbb R^2\to\Bbb R$$ is continuous and injective, and consider the subsets $$K_a=\{a\}\times[0,1]$$ of $$\Bbb R^2$$ for $$a\in\Bbb R$$. Each of these sets is compact and connected, and continuous functions preserve compactness and connectedness, so $$f[K_a]$$ is a compact, connected, subset of $$\Bbb R$$ for each $$a\in\Bbb R$$. The compact, connected subsets of $$\Bbb R$$ are precisely the closed intervals, so for each $$a\in\Bbb R$$ there are $$u_a,v_a\in\Bbb R$$ such that $$u_a\le v_a$$ and $$f[K_a]=[u_a,v_a]$$.

Finally, $$f$$ is injective, so the intervals $$[u_a,v_a]$$ are non-degenerate and pairwise disjoint. Thus, for each $$a\in\Bbb R$$ there is a $$q_a\in\Bbb Q\cap(u_a,v_a)$$, and since these intervals are pairwise disjoint, the map $$\Bbb R\to\Bbb Q:a\mapsto q_a$$ must be injective. But this is impossible, since $$\Bbb R$$ is uncountable while $$\Bbb Q$$ is countable, so there is no such function $$f$$.

• I cannot see why falsifying (a) automatically falsifies (b). For each space $X \ne \emptyset$ there are uncountably many continuous $f : X \to \mathbb R$, but if $X$ is countable or compact, there is no continuous surjection. Thus we need some additional information about $X$. Commented Dec 27, 2020 at 10:33
• @PaulFrost: Because I literally did not see the word surjective in (b). I’ll have to revise that part. Commented Dec 27, 2020 at 19:21
• @BrianM.Scott sir I think in the $2$nd paragraph $u_a \le v_a$ and $f[K_a]= [u_a,v_a]$ instead of $K_a =[u_a,v_a]$ Commented Jan 30, 2023 at 15:49
• @jasmine: You are absolutely right; thank you very much for catching that. Commented Jan 30, 2023 at 19:54