Let $(X,d)$ be a non-compact, complete metric space and $K\subseteq X$ compact. Pick $x^\* \in X\setminus K$.


Let $E$ be the connected component of $X\setminus K$ that contains $x^\*$. Let ${\cal C}$ be the collection of connected components of $K$. For each $C\in {\cal C}$ let $E\_C$ be the connected component of $X\setminus K$ that contains $x^\*$.


Do we have $E=\bigcap\{E\_C: C\in{\cal C}\}$?



Intersection of complements of connected components (2)
No.


[I'm assuming you means to write "let $E\_C$ be the connected component of $X \setminus C$ that contains $x^\*$." Otherwise I don't see how your question makes sense. Let me know if I'm guessing wrong and you meant to ask something else.]


Let $X\_0 = (\omega+1) \times \mathbb{R}$ (where $\omega+1$ has its usual topology as a convergent sequence). Let $X\_1$ be obtained from $X\_0$ by adding two points at infinity, $p$ and $q$. For a neighborhood basis for $p$, take sets of the form
$$(\omega+1) \times (-\infty,r) \cup \{p\}.$$
Similarly, for a basis of neighborhoods of $q$, take sets of the form
$$(\omega+1) \times (r,\infty) \cup \{q\}.$$
Let $X$ be obtained from $X\_1$ by adding infinitely many isolated points off to the side (these won't play any role in our argument, but you wanted $X$ to be not compact).


Let $K = \{p,q\}$ and let $x^\*$ be the point, say $(0,0)$ (although any other point of $X\_1 \setminus K$ will do just as well). Clearly $E = \{0\} \times \mathbb{R}$.


$K$ has two connected components, $\{p\}$ and $\{q\}$. Each of $X\_1 \setminus \{p\}$ and $X\_1 \setminus \{q\}$ is connected, so $E\_{\{p\}} \cap E\_{\{q\}} = X\_1 \setminus K = X\_0 \neq E$.


