Showing a set is open in a topology defined by invariant metric

Question

Let $U\subset\mathbb{C}$ be an open domain and $H(U)$ the space of holomorphic functions on $U$. Define the invariant metric $d(f,g)=\sum_{m=1}^{\infty}\frac{1}{2^{m}}\frac{\rho_{K_{m}}(f-g)}{1+\rho_{K_{m}}(f-g)}$, where $\rho_{K_{m}}(f)=\sup_{z\in K_m}|f(z)|$ and $\{K_m\}_{m\in\mathbb{N}}$ is a compact exhaustion of $U$. Prove that for $K\subset {U}$ compact, the set $B_K=\{f:\rho_K(f)<1\}$ is open in the metric induced by $d$.

My attempt: I let $f\in B_K$ and $f_n\to f$ in the metric ($d(f_n,f)\to 0)$, and tried to show that I have some $N$ s.t for every $n\geq N$, $f_n\in B_K$, but I wasn't able to do that. Also I wasn't able to show that I can find a $d$ open neighborhood of $f$ s.t $B_{r}(f)\subset B_K$. All I can say is that I have some $m\in\mathbb{N}$ s.t $K\subset K_m$.

Any hint would be appreciated.

Edited attempt: One can show that convergence on compacta is equivalent to convergence in this metric. Is this enough to conclude the requested result? Does this imply equivalence of topologies?

Answer 1

• Here by a compact exhaustion I mean the sequence $\{K_m\}_m$ of compact sets such that $K_n\subset \mathrm{int}K_{n+1}$ for all $n$ and that $\bigcup_mK_m=U$.
• Since $K\subset U$ is compact and $\{K_m\}_m$ is a compact exhaustion of $U$ then $K\subset K_p$ for some $p\in\Bbb N$.
• Fix $f\in B_K$. Put $\varepsilon = 1-\rho_K(f)$.
• Consider any $g\in B(f,r)$ where $$r=\frac 1{2^p}\cdot \frac\varepsilon{1+\varepsilon}.$$
• Then $$r>d(f,g)\geq \frac 1{2^p}\frac{\rho_{K_p}(g-f)}{1+\rho_{K_p}(g-f)}.$$
• From the monotonicity of $x\mapsto \frac x{1+x}$ we get $$\rho_K(g-f)\leq \rho_{K_p}(g-f)<\varepsilon.$$
• Finally, $\rho_K(g)\leq \rho_K(g-f) + \rho_K(f)<1$, i.e. $g\in B_K$, which shows that $B(f,r)\subset B_K$.