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

Question

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...

Please give hint. Thank you.

Answer 1

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 $f$.

Answer 2

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 $f$.

Answer 3

My method:

Recall the Intermediate Value Theorem: If $ f : X → \mathbb R $ is a continuous function from a connected metric space $X$ and $y ∈ (\inf f (X ), \sup f (X ))$, then there exists $x ∈ X$ such that$ f(x) = y$. [If $f(X)$ is not bounded below then we take $ \inf f(X) = −\infty $ and similarly if $ f(X)$ is not bounded above.]

Prove by contradiction. Suppose $ f : \mathbb R^2 → \mathbb R $ is a continuous injection. We draw a right triangle $ABC$ with $ A $ as a right angle in $\Bbb R^2$, $f $ maps this triangle to $\Bbb R$, we denote $X=AB$, $Y=AC$ $\cup$$BC$, $ X$ and $Y $ are both connected (since we can prove that they are both homeomorphic to closed interval, which is connected).

Since $f$ is injective, we have $f(A)\neq f(B)$, WLOG, $f(A)<f(B)$, then the interval $[f(A), f(B)]\subseteq [\inf f(X), \sup f(X)]\subseteq f(X)$($X$ is closed), $[f(A), f(B)]\subseteq$$[\inf f(Y), \sup f(Y)]\subseteq$ $f(Y)$($Y$ is closed).

Also, we have $X$$\cap$$Y$={$A$,$B$}, take $y=(f(A)+f(B))/2$, $y$ $\in$ $f(X)$ $\cap$ $f(Y)$={$f(A)$,$f(B)$}, but $y$$\neq$$f(A)$ or $f(B)$ as $ f$ is injective. We get contradiction.