Showing that arcs do not separate the plane $\mathbb{R}^2$ Is $\mathbb{R}^2\setminus f([0,1])$ connected if $f:[0,1]\to\mathbb{R}^2$ is an embedding?
It seems that this is clearly true but I am having a hard time proving it.
The only things that I know is that $f([0,1])$ is compact and simply connected. 
Since $H=\mathbb{R}^2\setminus f([0,1])$ is open in $\mathbb{R}^2$, we can prove that $H$ is locally path-connected and therefore its connected components and path-connected components are the same, or more precisely $A\subset H$ is a connected component if and only if is a path-connected component.  
Since $f([0,1])$ is bounded, it is contained in a disk $D$. Hence if $H$ has a separation $\{U,V\}$ (both non-empty, open and disjoint with union $H$), then $U\cap (\mathbb{R^2}\setminus D)$ and $V\cap (\mathbb{R^2}\setminus D)$ are open, disjoint with union $\mathbb{R^2}\setminus D$. Thus we can assume without lose of generality that $U\subset D$. In fact, either $U\cap (\mathbb{R^2}\setminus D)$ or $V\cap (\mathbb{R^2}\setminus D)$) is empty because $\mathbb{R^2}\setminus D$ is connected.
Finally we have that $\{U,V,f([0,1])\}$ is a partition of $\mathbb{R}^2$ where $U$ and $V$ are open and $f([0,1])=\partial U \cup \partial V$ and $U$ is bounded.
It sounds like I have to use the Jordan Curve Theorem somehow.
 A: I think you are right when you say you can use the Jordan Curve Theorem. First, we want to create a close curve, so we take a tubular neighborhood of the image of $f$ and we draw a parallel copy of $f$ in this tubular neighborhood. You can then join the end points of both segments by a straight line in your tubular neighborhood.
Then, by the Jordan Curve Theorem, this curve separates you plane in two connected region $\mathcal{U},$ $\mathcal{V}$. Moreover, you can prove that by construction, they will both be locally path connected and hence path connected. Therefore, you only need to prove that you can join an element from $\mathcal{U}$ to an element of $\mathcal{V}$ by a path that do not cross $f([0,1]).$ In order to do that, take a small neighborhood of the parallel copy of $f$ that you drew and take you line to be there. You then prove that $\mathbb{R}^2 \backslash f([0,1])$ is path connected, and hence connected
A: I was reading a little and I found out that this problem is far from being trivial, and  actually can be used as the core of the proof of the Jordan curve theorem. I am going to post some links that I found in the web with five different proofs of the Jordan arc theorem (which states that $\mathbb{R}^2\setminus f([0,1])$ is connected if $f$ is and embedding) and the Jordan Curve Theorem, respectively. One of them using the Janizewski's theorem,  other using the Brouwer's fixed-point theorem in which shows that if $\mathbb{R}^2\setminus f([0,1])$ is not connected then we can build a retraction from the disk $D^2$ to the circumference $S^1$, which is impossible. 
http://luisto.fi/documents/JordanCurveTheorem.pdf
Also I found an interesting proof using gratings:
http://pages.vassar.edu/mccleary/files/2011/04/FinalChapter9.pdf
This is another approach using standard paths and standard curves:
http://www.maths.ed.ac.uk/~aar/jordan/dosttind.pdf
And here using graphs.
But I am still interested in a proof using the Jordan Curve Theorem. @djvyu72 gave an idea of how we can apply it but still is incomplete.
A: You can use the result to prove the Jordan curve theorem (for example, see Fulton's book on Algebraic Topology).
Suppose, for the sake of a contradiction, that $I$ is homeomorphic to $[0,1]$ and $U\subseteq \Bbb R^2-I$ has two connected components. Divide $[0,1]$ into two halves intersecting at a point, and let $I_1$ and $I_2$ be the corresponding arcs. Then I claim that if $x,y$ are in different connected components of $U$, they must be in either different connected components of $U_1=\Bbb R^2-I_1$ or $U_2=\Bbb R_2-I_2$. We know there exists a locally constant function $f:U\to \Bbb R$ such that $f(x)\neq f(y)$. I claim that $f$ is of the form $f_1-f_2$ where $F_i:U_i\to\Bbb R$ is continuous locally constant and $f_i=F_i\mid U_1\cap U_2$. Indeed, we can construct a partition of unity $\psi_1,\psi_2$ subordinate to the open cover $U_1,U_2$ of $U_1\cup U_2$ and consider $F_1=f\psi_1$, $F_2=-f\psi_2$. 
Knowing this, the claim follows, for if $x,y$ were in the same connected components of both $I_i$, we could write $f=f_1-f_2$ and we would have $f_i(x)=f_i(y)$ contradicting our choice of $f$. 
We can continue this, to obtain a sequence of intervals $I_1,I_2,I_3,\ldots$ that halve in length at every step for which $x,y$ are in distinct connected components of each $\Bbb R^2-I_i$. Now $\bigcap I_i=\{p\}$ is a point and $\Bbb R^2-p$ is path connected, so we can take a path $x\to y$, and we may take a ball $B$ such that $p\in B$ doesn't intersect this path. But $\Bbb R^2-B$ is also path connected, and we can take $J$ big enough so that $I_J\subset B$, meaning that $\Bbb R^2-B\subseteq \Bbb R^2-I_J$. This contradicts the fact that $x,y$ were in different connected components, and finishes the proof. 
