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**Stiwan** I don't really see why all we need to make the norm continuous is "neighborhoods centered at 0". How would you prove that the norm is continuous in this case? Let's look at a special case to make things easier:

Suppose $\displaystyle (X,\|.\|) = (\mathbb{R},|.|)$. Then, the preimage of $\displaystyle B_r(x) := \{y \in R : |x-y|< r\}$ is equal to $\displaystyle B_r(x) \cup B_r(-x)$. How would you show that any set of this type is equal to a union of sets that are elements of your base?

Anyway, I think I've found quite an easy way to prove it: In general, it holds that the preimage of any subset of $\displaystyle [0,\infty)$ with respect to $\displaystyle \|.\|$ is symmetric. Because the preimages of open subsets of $\displaystyle [0,\infty)$ form a subbase of the initial topology and the union and intersection of symmetric sets is always a symmetric set itself, any set that is open in the initial topology must be symmetric. As $B_r(x) = \{y \in X: \|x - y\| < r\}$ is not symmetric if $ x \neq 0$ but open in the topology induced by the norm, the topologies can not be equal to each other.