Assume $ U $ is an open subset, and assume $ D=D(a,r) $ is a disk such that $\overline{D} \subset U$ (The closure of D contained in U ). Can we find $ \varepsilon>0 $ such that $D(a,r+\varepsilon) \subseteq U $ ? (an open disk with radius $r+\varepsilon$ which obviously contains D )
Furthermore, can we even find such disk, that even its closure contained in $ U$ ?
If so, how can we prove it?
I started to learn complex analysis and it seems like the author using topology results without a proof, and I havent learned topology yet, so im curious about this one.
Thanks in advance.
Best Answer
This is true because, in the metric space of complex numbers, bounded closed sets, like the boundary of a closed disk, enjoy a useful topological property called compactness. A compact subset $X$ has the property that if you have $X \subseteq \bigcup_i O_i$ where the $O_i$ are a collection of open sets (e.g., open disks), then there is a finite subcollection of the $O_i$, say $O_{i_1}, \ldots, O_{i_n}$ such that $X \subseteq O_{i_1} \cup \ldots \cup O_{i_n}$. In your example, you can pick a small open disk $D_x$ around each $x$ in the boundary $\delta(\overline{D})$ of $\overline{D}$ such that $D_x \subseteq U$. You then have that $\delta(\overline{D}) \subseteq \bigcup_x D_x$ and hence, by compactness, $\delta(\overline{D}) \subseteq D_{x_1} \cup \ldots \cup D_{x_n}$ for some finite set $x_i$ of elements of $\delta(\overline{D})$. If you now take $\varepsilon$ to be the minimum of the radii of the disks $D_{x_1}, \ldots, D_{x_n}$, you will have $D(a, r + \varepsilon) \subseteq U$.