The assertion is clear if $e_{1}(x)=0$. For $e_{1}(x)\neq 0$, let $e_{n}(x)=e_{1}^{\otimes n}(x)$, note that

$$\lvert e_{1}^{\vee}(x)\rvert_{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=(\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}})^{\frac{1}{n}}.$$

By Corollary 3.15, one has

$$\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}$$

so

$$\lvert e_{1}^{\vee}(x)\rvert_{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}^{\frac{1}{n}}.$$

Tautologically, one has

$$s_{n,j}(z)=(e_{1}^{\otimes n})^{\vee}(x)(s_{n,j})=e_{n}^{\vee}(x)(s_{n,j}),$$

so the criterion holds. By these defining equations, it is easy to see that the image of the open dual unit disc bundle is an open set. ∎