Is this topology on the space of locally bounded functions metrizable?

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Let

  • $(M,d)$ be a metric $\mathbb R$-vector space
  • $\overline B_\varepsilon(x)$ denote the closed ball with radius $\varepsilon>0$ around $x\in M$
  • $E$ be a $\mathbb R$-Banach space

Now, let $$\left\|f\right\|_{B(K,\:E)}:=\sup_{x\in K}\left\|f(x)\right\|_E\;\;\;\text{for }f:M\to E$$ for $K\subseteq M$ and $$B_{\text{loc}}(M,E):=\left\{f:M\to E\mid\left\|f\right\|_{B(K,\:E)}<\infty\text{ for all compact }K\subseteq M\right\}.$$

Let $\tau$ denote the topology on $B_{\text{loc}}(M,E)$ induced by $\left\{\left\|\;\cdot\;\right\|_{B(K,\:E)}:K\subseteq M\text{ is compact}\right\}$.

Question 1: Is $\tau$ equal to the topology induced by $\left\{\left\|\;\cdot\;\right\|_{B(\overline B_n(0),\:E)}:n\in\mathbb N\right\}$?

Question 2: If the answer to question 1 is yes, are we able to conclude that $\tau$ is metrizable via $$d(f,g):=\sum_{n\in\mathbb N}\frac1{2^n}\left\|f-g\right\|_{B(\overline B_n(0),\:E)}\;\;\;\text{for }f,g\in B_{\text{loc}}(M,E)?$$