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Let $\mathbb{D}$ denote the unit disc. Say $U\subset \mathbb{D}$ is an open set such that for every $r\in [0,1)$ there exists $z\in U$ such that $|z| = r$. Given $0<\rho<1$ is it then possible to find a compact subset of $U$ denoted $K$ such that $\forall r\in[0,\rho]$ there exists $z\in K$ such that $|z| = r$?

If $U$ is connected then this is trivially true since $U$ is pathwise connected and we need only pick a point $\zeta\in U$ with $|\zeta| = \rho$ and then connect it via a curve to $0$ and consider the trace of the curve as our compact set. A similar construction is valid if $U$ consists of finitely many components. However if $U$ consists of infinitely many components is it still possible to find such a compact set?

OgvRubin
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1 Answers1

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The solution was more direct than I thought:

For every $r\in [0,\rho]$ we can find $z_r\in U$ such that $|z_r| = r$. Since $U$ is open we can find $\epsilon_r>0$ such that the closure of $B_r := B(z_r,\epsilon_r)$ is contained in $U$. Now let $T:\mathbb{D}\rightarrow[0,1)$ be defined by $T(z) = |z|$. Then $(T(B_r))_{r\in [0,\rho]}$ is an open cover of the compact set $[0,\rho]$ and therefore only finitely many are needed to cover $[0,\rho]$: say $[0,\rho]\subset \cup_{k=1}^{n}T(B_{r_k})$. Defining $K$ by

$$K = \left(\bigcup_{k=1}^{n}\overline{B_{r_k}}\right)\cap \overline{B(0,\rho)}$$

we see that $K$ solves our problem.

I guess this generalizes to the following situation. Let $\Omega$ and $U$ be metric spaces where closed and bounded sets are compact and let $T:\Omega\rightarrow U$ be an open surjective map. Then if $C$ is a compact subset of $U$ we can find a compact subset of $\Omega$ denoted $K$ such that $T(K) = C$.

OgvRubin
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