Find the number of relatively compact connected components Let $X=\{(x,y):x^2+y^2<5\}$ and $K=\{(x,y):1\leq x^2+y^2\leq 2 or 3\leq x^2+y^2\leq 4\}$.  Then:
Find the number of relatively compact connected components of $X\setminus K$ in $X$.
I drew a picture of the problem and I found since the boundary i.e $x^2+y^2=1 or x^2+y^2=2$ gets erased in $X\setminus K$ so it has zero connected components .
But the answer given is $2$ .Why?Any help

You have $$X \setminus K = D \cup E \cup F$$ where $D=\{(x,y):x^2+y^2 < 1\}$, $E=\{(x,y): 2 < x^2+y^2 < 3\}$ and $F=\{(x,y): 4 < x^2+y^2 < 5\}$.
$D$, $E$ and $F$ are connected components of $X$. $D$ and $E$ are also relatively compact in $X$. But not $F$.
So the answer is indeed $2$.
