Bounded connected component of $\mathbb{R}^n\setminus f(\partial \varOmega) $ is contained in any ball containing $f(\bar \varOmega)$

Question

I'm reading the book: "Deimling Nonlinear Analysis" and I stumbled upon something I can't really figure out, at page 24, when proving the product formula of the topological degree of a composition.

In the proof it's made the following statement: Let $\varOmega$ be an open set, $f\in C^1(\varOmega)\cap C^0(\bar\varOmega)$ and $K_i$ a bounded connected component of $\mathbb{R}^n\setminus f(\partial \varOmega)$. If $f(\bar \varOmega)\subset B_r(0)$ (the euclidean ball), then $K_i\subset B_r(0)$

Now, I tried to see if I could conclude anything by noting that $f(\partial \varOmega)\subset f(\bar \varOmega)\subset B_r(0) $, but I can't get anything useful. So I'm asking if anyone has a proof for it. Also, I know that if I state that $K_i$ is bounded I can chose an appropriate radius that does that, but it seems to me that in this proof it's stated something stronger.

Answer 1

$\Bbb{R}^d \setminus B_r (0)$ is connected (if $d > 1$, otherwise it is the union of two unbounded, connected sets) and contained in $\Bbb{R}^d \setminus f(\partial \Omega)$. Thus, if a connected component of $\Bbb{R}^d \setminus f(\partial \Omega)$ intersects $\Bbb{R}^d \setminus B_r (0)$ nontrivially, it needs to contain an unbounded component, and can thus not be bounded.

Hence, if there exists a bounded component of $\Bbb{R}^d \setminus f(\partial \Omega)$, it needs to be contained in $B_r(0)$.