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I'm trying to do the following exercise from my lecture notes:

There does not exist a continuous injection from $\mathbb{R}^n$ to $\mathbb{R}$, for all $n \geq 2$.

I don't really know where to start, but I think it helps that the exercise is marked as harder than the following exercises before it:

  • $\mathbb{R}^n\setminus\{0\}$ is connected, for all $n \geq 2$.

  • $\mathbb{R}^n$ is nonhomeomorphic to $\mathbb{R}$, for all $n \geq 2$.

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  • @DietrichBurde: The existence of a continuous injection $\mathbb{R}^n \to \mathbb{R}$ is very different from an arbitrary injection. – anomaly Jun 01 '16 at 19:30
  • Yes, right - I had only read the comment "There is such an injective function, but not a continuous one." This is standard, I was searching for it on MSE. – Dietrich Burde Jun 01 '16 at 19:31
  • Let $f \colon \mathbb{R}^n \to \mathbb{R}$ be continuous. Let $B = { x \in \mathbb{R}^n : \lVert x\rVert \leqslant 1}$. Consider $f(B)$, and note that there are many paths in $B$ connecting any two given points. – Daniel Fischer Jun 01 '16 at 19:32
  • A sledgehammer, but nonetheless illuminating, result would be the following - https://en.wikipedia.org/wiki/Invariance_of_domain – Hmm. Jun 01 '16 at 19:33
  • Now this question answers the question. The answer of Pete Clark is for all $n\ge 2$. – Dietrich Burde Jun 01 '16 at 19:35
  • Given such a map $i:\mathbb{R}^n \to \mathbb{R}$, consider the connected space $i(\mathbb{R}^n\setminus {p}) = i(\mathbb{R}^n)\setminus {i(p)}$. – anomaly Jun 01 '16 at 19:36
  • @anomaly, that won't help unless you know that image of $ \mathbb{R}^n$ is open, which it actually is in this case. – Hmm. Jun 01 '16 at 19:40

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