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I wish to prove that no continuous, one-to-one, and onto function from $\Bbb R$ to $\Bbb R^2$ has a continuous inverse.

I have some ideas that seem relevant:

  1. Continuous functions preserve connectedness, and taking a point out of $\Bbb R$ makes it disconnected, while taking a point out of $\Bbb R^2$ leaves it connected
  2. $\Bbb R$ is "smaller" in some sense than $\Bbb R^2$, since $\Bbb R$ is a subset of $\Bbb R^2$

By $\Bbb R$ I mean the reals and by $\Bbb R^2$ I mean the Cartesian product of the reals with the reals.

Izaak van Dongen
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Memento Mori
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  • Welcome to MSE! Your first idea sounds very good to me. What problem do you run into when trying to make a proof out of that? – Izaak van Dongen Oct 30 '23 at 23:16
  • Thanks! I don't understand why I can take a point out of R/R^2 and still have it be a generalized proof about any possible, continuous bijective function – Memento Mori Oct 30 '23 at 23:18
  • Alright. I'll start you off one how to write the proof. "Suppose $f: \Bbb R \to \Bbb R^2$ is a continuous bijection with continuous inverse. Consider the restriction of $f$ to $\Bbb R \setminus {0}$, and the set $\Bbb R^2 \setminus {f(0)}$. This restriction of $f$ has some properties...." – Izaak van Dongen Oct 30 '23 at 23:20
  • Your idea (1) is a nice idea, but you have to focus on continuity of the inverse function. – Ted Shifrin Oct 30 '23 at 23:25
  • A continuous function with continuous inverse is called a homeomorphism. If there is a homeomorphism between two spaces, we call them homeomorphic. Your question is the same as "how to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic". – Jakobian Oct 30 '23 at 23:35

1 Answers1

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Suppose such a function exists, i.e., a bijective function $f: \Bbb R \to \Bbb R^2$ with a continuous inverse. By translating, i.e., defining $g(x) = f(x) - f(0)$, we have another continuous function $g$ that is one-one, onto, with a continuous inverse. This is easy to check;

  1. Suppose $g(x) = g(y)$. Then, $f(x) = f(y)$ is immediate. We have $x=y$.
  2. Let $(p,q) \in \Bbb R^2$. Consider the point $f(0) + (p,q) \in \Bbb R^2$. As $f$ is onto, there exists $x\in \Bbb R$ with $f(x) = f(0) + (p,q)$. Thus, $g(x) = (p,q)$.
  3. As $g$ is bijective, its inverse exists. Moreover, $g^{-1}(p,q) = f^{-1}(f(0) + (p,q))$. As $f^{-1}$ is continuous, so is $g^{-1}$.

In other words, it is safe to assume $f(0) = (0,0)$.

As you suggest, let's consider the restriction map $f: \Bbb R\setminus\{0\} \to \Bbb R^{2}\setminus\{(0,0)\}$. This restriction map is one-one, onto, and continuous with a continuous inverse. $\Bbb R^{2}\setminus\{(0,0)\}$ is connected, but $\Bbb R\setminus\{0\}$ is not. As connectedness is preserved by homeomorphisms, this is the desired contradiction.

stoic-santiago
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