Counterexample in connected set I'm asked to find an example of two connected sets $X$ and $Y$, $X\subset Y$ and $C$ a component of $Y\setminus X$ such that $X\cup C$ is not connected.
I figured $X$ must not be closed because then $X\cap C\neq \emptyset$ and $C\cup X$ would be connected. But that's all I got. Any hints would be appreciated.
 A: Maybe overkill, but the Kuratowski fan fits the bill.
Taking $Y=$fan and $X=\{p\}$ the troublesome point (which, since it is just a point, is connected), we have that $Y- X$ is totally disconnected. Therefore, take any other point $C=\{q\}, q \in Y-X$ as a connected component. It follows that $X \cup C$ is not connected (being a two-point subspace of $\mathbb{R}^2$).
A: $$Y=\big([0,1]\times \{0\}\big)\cup \{\langle 0,1\rangle\}\cup \bigcup_{n\geq 1} \{1/n\}\times [0,1]$$
$$X=[0,1]\times \{0\}$$
$$C=\{\langle 0,1\rangle\}$$
It is impossible to find an example if $Y$ is compact Hausdorff.  
Also, $Y\setminus C$ is always connected.  
A: Whether or not $X$ is closed, $X \cap C$ is empty. It is different with $C$,
which is closed in $Y \setminus X$ (being a component). It follows that $\overline{C} \cap Y \setminus X = C$ (closure taken in $Y$). Also $\overline{C} \cap X$ must be empty (otherwise $X \cup C$ would be connected). Hence $\overline{C} = C$, meaning $C$ is closed in $Y$. Pick a point in $Y \setminus (X \cup C)$, take the component of $Y \setminus X$ at that point and repeat the argument. With a finite number of components $C_1, .. ,C_n$ available, all closed in $Y$, we still have a disconnected subspace $X \cup C_1 \cup .. \cup C_n$ which therefore cannot be the whole of $Y$. This delivers a novel point in $Y \setminus X$ and another component $C_{n+1}$ and we can continue the argument ad infinitum.
So at least one can see that the required example must have an infinitely fragmented $Y \setminus X$.
