Let $U\subseteq \mathbb C$ be open and connected. If $f\colon [0,1]\to U$ is continuous with $f(0)\neq f(1)$ and $f(s)\neq f(t)$ for $s\neq t$, then $U\setminus f([0,1])$ is connected.

This seems to be "obviously true", so I'm looking for a proof. What I have so far:

As $U$ is open, it is path-connected. I want to show, that $V:=U\setminus f([0,1])$ is path-connected as well. Let $a,b\in U$, then we have $\varphi\colon[0,1]\to U$ continuous with $\varphi(0)=a$ and $\varphi(1)=b$. If $\varphi([0,1])\subset V$ we're done. Now assume that $\varphi([0,1])\notin V$.

For sufficient small $\varepsilon>0$ we have $0=t_0<\dots <t_n=1$ with $$f([t_{j-1},t_j])\in K_{\varepsilon}(f(t_j)\subset U,~j=1,\dots n$$ where $K_{\varepsilon}(f(t_j))$ denotes the open disk with radius $\varepsilon$ and origin $f(t_j)$.

I want to cover the curve $f$ with small, overlapping circles, so when $\varphi$ gets near $f$, it first has to cross over $\partial K_{\varepsilon}(f(t_i))$ and I can use the $\partial K_{\varepsilon}(f(t_i))$ to get "around $f$".

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I have proven, that such circles as above exist, but do they always overlap?

Edit: for my application it is enough to further assume that $f$ is differentiable. This might help/I might have an idea that I'll think through.

  • $\begingroup$ I think (just intuitively) that these epsilons, they won't all be the same. You may need to vary them based on some local properties of the curve. $\endgroup$ – peter.petrov Jul 16 '15 at 15:21
  • $\begingroup$ @peter.petrov, I have proof that they are. ;) Edit: not the same for every curve of course, but for a fixed curve I find such epsilon as described. $\endgroup$ – Hirshy Jul 16 '15 at 15:22
  • $\begingroup$ OK, sorry ;) was just thinking out loud. $\endgroup$ – peter.petrov Jul 16 '15 at 15:29
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    $\begingroup$ This is the "Jordan arc theorem", and if you search that term on math.stackexchange and on mathoverflow you'll find closely related questions. However, most of those questions prove something stronger and/or use big cannons. So I still think the best proof---meaning simplest, using the least technology---is in Munkres topology book (theorem 13.3 in the old version of his book). Nonetheless, the proof in Munkres is not as straightforward as what you are suggesting, in particular it uses a special version of the Van Kampen theorem. $\endgroup$ – Lee Mosher Jul 16 '15 at 16:00
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    $\begingroup$ Yes, you said it's not a closed curve. You need to think about the case where the curve is closed anyway; if you think you've converted the above into an actual proof a good check is whether the same proof would work for a closed curve, in which case the proof is wrong. In particular, when you use those circles to get around $f$ there has to be some point in the argument that uses the fact that $f$ is not closed. $\endgroup$ – David C. Ullrich Jul 16 '15 at 16:01

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