Given a closed disc staying in an open set in $\mathbb{C}$, is there a larger open disc within the open set which contains this closed disc?

Question

Let $U$ be open and $D \subset U$ be a closed disc. In the book Complex Analysis by Stein (P.39 Corollary 2.3), he claimed in the proof that there exists a slighly larger open disc $D' \subset U$ such that it contains $D$.

However, I doubt the validity of this statement. Say if I denote $d(x,\partial U)$ to be the distance of the point $x$ from the boundary of $U$, maybe there will be some open set $U$ such that $D \subset U$, yet there exists some sequence of points $x_n \in \partial D$ such that $d(x_n,\partial U)=\frac{1}{n}$. (I am not sure if such open set exists.)

Intuitively (maybe wrong), there is no way for us to draw a open disc $D'\subset U$ such that $D \subset D'.$

Is Stein doing wrong or my construction is faulty? Thank you.

Answer 1

It's essentially a compactness result. There are quite a few ways you could look at this, but here's one way:

Suppose there was no such larger disk in $U$, enveloping $D = B[z; r]$. Then, $B\left[z; r + \frac{1}{n}\right] \setminus U \neq \emptyset.$ That is, we have a sequence $w_n \in B\left[z; r + \frac{1}{n}\right] \setminus U$ for all $n$. Such a sequence satisfies $|z - w_n| \le r + \frac{1}{n} \to r$, and $w_n \notin U$. Such a sequence is bounded, and hence has a convergent subsequence $(w_{n_k})$ converging to some $w$. Since $\mathbb{C} \setminus U$ is closed, we have $w \notin U$. But, at the same time we have $|z - w_{n_k}| \to |z - w|$, hence $|z - w| = r$ by uniqueness of limits. Thus, we have a point in $\mathbb{C} \setminus U$, but that is also in $D$. This is a contradiction.

Answer 2

Define a function $f(z)=d(z,\mathbb C\setminus U).$ Since $f$ is a continuous function and $D$ is a compact set, $f$ has a minimum value on $D,$ call it $m.$ Since $U$ is open, $f(z)\gt0$ for all $z\in U;$ since $D\subset U,$ it follows that $f(z)\gt0$ for all $z\in D;$ in particular $m\gt0.$ If $D$ has center $z_0$ and radius $r,$ then the open disc with center $z_0$ and radius $r+\frac m2$ contains $D$ and is contained in $U.$