Can we find $\lambda$ with $B(\lambda, |\lambda|) \subseteq K?$ Let $K$ be a compact subset of the positive real axis, viewed as a subset of the complex plane $\mathbb{C}$.
Let $a$ be an element not on the positive real axis, i.e. $a \in \mathbb{C}\setminus [0, \infty[$.
In a proof I'm reading, it is claimed that there is $\lambda \in \mathbb{C}$ with
$$K\subseteq B_\mathbb{C}(\lambda, |\lambda|), \quad a \notin B_\mathbb{C}(\lambda, |\lambda|)$$
This is intuitively clear but I can't formally prove it.
This may have something to do with separations axioms or Lebesgue coverings. 
 A: Let $x_K = \sup K$ and without loss of generality assume that $a$ lies in the first quadrant. Note that the boundary $C$ of the ball $B_{\Bbb{C}}(\lambda, \lvert \lambda \rvert)$ is just a circle with radius $\lvert \lambda \rvert$, passing through $0$. Therefore, we're looking for a circle like in the picture below, where $c = \lambda$, and where $x_C$ is the other intersection of $C$ with $\Bbb{R}$.

Clearly (by construction) the distances between $c$ and $O$, $c$ and $x_C$, and $c$ and the highest point of $C$ are all $\lvert \lambda \rvert$. Therefore such a circle exists if and only if
$$
\begin{cases}
x_C = 2\cos(\alpha) \lvert \lambda \rvert \geq x_K \\
\lvert \lambda \rvert < \sin(\alpha) \lvert \lambda \rvert + \operatorname{Im}(a)
\end{cases}
$$
where $\alpha = \lvert \arg(\lambda) \rvert$ is the angle between the real axis and the segment joining $\lambda$ and $0$. In particular, these equations have a solution if and only if
$$
\frac{x_K}{\operatorname{Im}(a)} < 2 \frac{\cos(\alpha)}{1 - \sin(\alpha)}
$$
which is always possible since $\frac{\cos(\alpha)}{1 - \sin(\alpha)} \to \infty$ as $\alpha \to \frac{\pi}{2}$.
A: Under the assumption that $a\leq 0$, let $s= \sup K <\infty$ and $t = \inf K$.
Because $K \subset \mathbb{R}^+, t > 0$ and we have that $K \subset B(s,s-t/2)$ and $a \notin B(s,s-t/2)$ because $ B(s,s-t/2) \subset (t/2, \infty) \subset \mathbb{R}^+$.
