A question about connected sets in $\mathbb{R}^2$

Let $C\subseteq\mathbb{R}^2$ be connected, open and have a bounded complement. Let $u\in C$ and $f:[0,1]\rightarrow \mathbb{R}^2$ be a continuous injective function such that $f(0)=u$. It is also given that $f([0,1))\subseteq C,f(1)\notin C$. Does it follow that $C-f([0,1])$ is connected ?

I really think that the answer is yes and the proof would be short, however I can't figure it out.

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Using stereographic projection, the question is closely related to math.stackexchange.com/questions/287062/… – Seirios Feb 16 '13 at 13:51
Like with Jordan curves properties, it seems obvious when one tries to draw an example. But for a short elementary proof? Let's see what all the clever people out there will find. – 1015 Feb 16 '13 at 14:50
By the way, where does the question come from? – 1015 Feb 16 '13 at 15:33
Editing the title to bump the question after two hours? Come on. – Chris Eagle Feb 16 '13 at 15:41
@Chris Eagle The reason why I edited it the title is because I did not get an answer. I wanted to inform MSE users that the question is still unsolved. I thought some people will not look at the question because it is solved. Now that there is a new answer I will remove the "still unsolved"from the title of the question – Amr Feb 16 '13 at 15:46

Since $C$ is open and connected, it is also path-connected. Let $C' = C \setminus \operatorname{img}(f)$. We prove that $C'$ too is path-connected.

Let $a, b ∈ C'$ and $γ \colon [0..1] → C$ be a path connecting those two points. If $\operatorname{img}(γ) ⊂ C'$ then there is nothing to show.

So let's assume $\operatorname{img}(γ) \not⊂ C'$, meaning $\operatorname{img}(γ) ∩ \operatorname{img}(f) ≠ ∅$. The intersection of those paths must be closed, so following $γ$ we eventually arrive at a first point of intersection (which isn't $a$) and at a last point of intersection (which isn't $b$).

Now since is $[0..1]$ is compact, $\operatorname{img}(f)$ is hausdorff, $f$-images of closed sets are indeed closed in the image of $f$ and since $f$ is injective, $\operatorname{img}(f)$ is indeed homeomorphic to $[0..1]$.

This implies that $\operatorname{img}(f)$ is locally connected, so we can cover it with a corresponding collection of open disks. We can find a Lebesgue number $r > 0$ such that any $r$-ball around a point in $\operatorname{img}(f)$ lies in the union of the covering.

Now their union yields some kind of inflation of the path $f$, so that it is open.

The idea is to jiggle $f$ around in this inflation such that it continues $γ$ to $a$ and $b$. But I fear I can't finish this argumentation, for one has to consider the boundary as well and I'm getting tired. So instead I made this beautiful picture in gimp using my trackpoint:

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It seems to me that this is essentially the same argument as ofer's. And like in ofer's argument, the end is not so obvious. – 1015 Feb 16 '13 at 15:48
@julien Yes, and I was pretty angry with him that he posted it before me. : - ( – k.stm Feb 16 '13 at 15:57
These things happen all the time over here... – 1015 Feb 16 '13 at 15:58
Don't forget the weird continuous functions – Amr Feb 17 '13 at 19:35

The complement of an open set $U$ in $\mathbb{R}^2$ consists of a union of simply connected sets iff $U$ is open. Notice that deleting the arc is the same as adding an arc to a closed set which is a component of the complement. But the closed set plus the arc clearly deformation retracts onto the original closed set, so it is simply connected iff the original is. So we see that adding the arc does not change the connectedness of the set.

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We can show that for every $x \in U \setminus f([0,1])$, we can get from $x$ to $u$ with a curve in $U \setminus f([0,1])$.

Let $g$ be the curve that connect between $x$ and $u$ in $U$. If $f$ and $g$ don't intersect, we're done.

Otherwise, let $y$ be the first point where the two curves intersect. $f$ is injective, therefore exists unique $0<t<1$ s.t. $y=f(t)$.

The set $[0,t]$ is compact in $U$ open set, therefore there exists $\delta>0$ s.t. $B(f([0,t]), \delta) \subset U$. It's easy to see that the set $B(f([0,t]), \delta) \setminus f((0,t))$ is connected. So there exists a curve $h$ from $f(t)=y$ to $f(0)=u$ and by concatenation of $h$ with the first part of $g$ which connect from $x$ to $y$, we'll get a curve in $U \setminus f([0,1])$ from $x$ to $u$.

Edit: I've mistakenly shown that $U \setminus f((0,1))$ is connected. So to fix it, for $x,y \in U \setminus f([0,1])$, do the same process as describe above for $t_0,t_1$ where $f(t_0), f(t_1)$ are the first and last intersections of the original curve between $x$ and $y$ and the curve $f$.

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But $u=f(0)$ is not in this set. – 1015 Feb 16 '13 at 14:40
You are right. Fixed it. – ofer Feb 16 '13 at 15:08
Is it really that easy to see that the set $B(f([0,t]),\delta)\setminus f((0,t))$ is connected? – 1015 Feb 16 '13 at 15:15
I agree with julien: If $T$ is a "tubular neighborhood" of the path $f$, then it is intuitively true that $T\setminus f\bigl([0,1]\bigr)$ is connected, but a rigorous proof is welcomed... – Matemáticos Chibchas Feb 16 '13 at 19:57

The complement $\mathbb{R}^2 \backslash C$ is a homeomorphic image of finitely many disjoint closed balls. Now you can apply Jordan arc theorem, and you are done.

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