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Let $V$ be an open subset of $\mathbb{R}^3$ (but since I don't think there's anything special about $3$ in this case, let's say $\mathbb{R}^p$). Prove that there can't exist a continuous injective function $f:\ V \rightarrow \mathbb{R}$

Here's what I've tried:

Assume that there exists such a functien $f$.

  • For every open $A \subseteq V:\ f(A)$ is open in $\mathbb{R}$
  • For every open $W \in \mathbb{R}:\ f^{-1}(W) $ is open in $V$ (definition of continuity).
  • Because $f$ is injective, $f^{-1}:\ f(V) \rightarrow V$ can be regarded as a function and is a bijection.

I don't see how any of these can be a problem.

I can't even see a problem intuitively. $\mathbb{R}^3$ is not somehow 'larger' than $\mathbb{R}$..

EDIT: This is not a duplicate of Question clarification., as that question was about what a sentence meant, not about answerring it.

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Syd Kerckhove
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1 Answers1

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Let $f:V \subseteq \mathbb{R}^p\rightarrow \mathbb{R}$ be a continuous injective function (with $p\ge 2$).

If $V$ is empty then there is no problem, there is a continuous injective function, the empty function.

If $V$ is non-empty, then there exists an open ball $B(x,\delta) \subseteq V$. This open ball is convex and thus connected. Because $f$ is continuous, the image of a connected set must be connected as well and must therefore be an interval in $\mathbb{R}$. Let $X$ be $B(x,\delta) \setminus \{x\}$. This is still connected ($p\ge 2$) so $f(X)$ must be an interval (connected) as well. Because $f$ is injective: $$ f(X) = B(x,\delta) \setminus f(\{x\} = f(B(x,\delta))) \setminus f(\{x\}) $$ ... but this is not an interval. Contradiction.

Syd Kerckhove
  • 705