Infimum of the supremum absolute value of a decreasing sequence of subsets of $\mathbb{C}$ with non-empty intersection

Question

Let $K_{n}$ be a decreasing sequence of bounded subsets of $\mathbb{C}$ such that $\cap_{n}K_{n}=K\neq\emptyset$. Let $\lambda_{n}=\text{sup }_{\lambda\in K_{n}}|\lambda|$ and $\lambda_{0}=\text{sup }_{\lambda\in K}|\lambda|$. Then $\{\lambda_{n}\}$ is a decreasing sequence of positive numbers bounded below by $\lambda_{0}$. Is it true that $\lambda_{0}=\text{inf }\lambda_{n}$? It seems like it should be true, I would be grateful for a hint on this!

Answer 1

In general, this is false. Let $K_n = \{0\}\cup (1-1/n,1 )\subset \mathbb{R}$, for example: then $\lambda_n=1$ for all $n$ but $\lambda_0=0$.

If $K_n$ are compact, this is true. Let $\lambda = \inf \lambda_n$ and consider the sets $A_n = K_n \cap \{z: |z|\ge \lambda\}$ which are nested, compact, and nonempty. Conclude that $K \cap \{z: |z|\ge \lambda\}$ is nonempty, hence $\lambda_0\ge \lambda$. The reverse inequality is clear.