Extending a connected open set Assume $\emptyset\neq V\subseteq U\subseteq\mathbb{R}^n$ are open and connected sets so that $U\setminus\overline{V}$ is connected as well. Given any point $x\in U$, is there always a connected open set $W\subseteq U$ so that $\{x\}\cup V\subseteq W$ and $U\setminus\overline{W}$ is connected? In other words, can $V$ be extended to a connected open set containing a given point so that the complement of the closure of the extended set is still connected?

There are two cases to consider:


*

*$n\ge 2$. Observe that in this case, for every open connected set $S\subset R^n$, any $a\in A$ and any sufficiently small $r\ge 0$, the complement of the closed ball 
$$
A\setminus \overline{B(a, r)}
$$
is still connected. Now, if $V$ is dense in $U$ then the only meaningful answer is to take $W=U$ (for any choice of $x$) and then $U\setminus \bar W$ and $W$ are both connected (the first one is empty of course). If $V$ is not dense in $U$, we can do a bit better than this: The subset $U\setminus \{x\}\cup \bar V$ is open and nonempty. Pick any point $a$ in this complementary set and let
$$
W= U \setminus \overline{B(a, r)}
$$
where $r>0$ is sufficiently small. Then 
$$
U\setminus \bar W = B(a, r)
$$
(the open ball) is connected and nonempty and $W$ is also connected by the above remark. 

*$n=1$ (I will leave out the case $n=0$). Then both $U$ and $V$ are intervals an you can take $W$ to be the smallest open interval containing $V$ and $x$.  
