A nonseparation theorem for arcs on $S^{2}$ Let $A$ be a simple unclosed curve in $S^{2}$. Is there a simple way to prove that $S^{2}\setminus A$ is path-connected using homology? 
By simple unclosed curve I mean that $A:I\rightarrow S^{2}$ is continuous and inyective.
 A: Let $A\subset S^2$ be a homeomorphic image of the closed interval. Then by Alexander duality
$$
\tilde{H}_0(S^2-A)\cong H^1(A)=0. 
$$
Therefore, $S^2 -A$ is (path) connected. 
A: Thanks to studiosus hint, I believe I've solved my problem. 
Let $U =S^{2}- A([0,\frac{1}{2}])$ and $V=S^{2}-A([\frac{1}{2},1])$. Given that $S^{2}$ is Hausdorff then both $U$ and $V$ are open in $S^{2}$. Note that $U\cap V = S^{2}-A(I)$, and $U\cup V = S^{2}-A(\frac{1}{2})\cong\mathbb{R}^{2}$. Therefore $\widetilde{H}_{i}(U\cup V)=0$ for all $i\geq0$. Next we consider the Mayer-Vietoris sequence:
$\cdots\rightarrow\widetilde{H}_{1}(U\cup V)\rightarrow\widetilde{H}_{0}(U\cap V)\rightarrow\widetilde{H}_{0}(U)\oplus\widetilde{H}_{0}(V)\rightarrow\widetilde{H}_{0}(U\cup V)\rightarrow\cdots$
which tells us that $\widetilde{H}_{0}(U\cap V)\cong\widetilde{H}_{0}(U)\oplus\widetilde{H}_{0}(V)$. If there were an element in $\widetilde{H}_{0}(U\cap V)$ different from $0$, then there'd be an elemento different from $0$ in either $\widetilde{H}_{0}(U)$ or $\widetilde{H}_{0}(V)$. 
Suppose there is such an element in $\widetilde{H}_{0}(U)$. Divide the interval $[0,\frac{1}{2}]$ in two (this will give two new open sets $U_{1}$ and $U_{2}$) and replicate the argument we have just done to show that there is a non trivial element in either $\widetilde{H}_{0}(U_{1})$ or $\widetilde{H}_{0}(U_{2})$. If one continues this process, one will get a sequence of closed nested intervals $\cdots I_{3}\subset I_{2}\subset I_{1}$ such that the length of $I_{n}$ is $\frac{1}{2^{n}}$, so that the only common point to all of them will be a point $p$. Because $S^{2}-\left\{p\right\}\cong\mathbb{R}^{2}$, it is path-connected, and one can trace back the homology groups and conclude that $\widetilde{H_{0}}(S^{2}-A)=0$.  
