Is the norm topology the same as the initial topology generated by the norm function? I have a question about the topology on a normed vector space. 
A normed vector space $(X, \| \cdot \|_X)$ naturally comes with a topology, known as the norm topology or strong topology on $X$, generated by the open balls $B (x, r) := \{y \in X : \|y-x\|_X < r\}$. Is this the same as the intial topology on $X$ generated by the norm function $\| \cdot \|_X : X \to \mathbb{R}$, i.e., the smallest topology that makes this norm function continuous?
I thought yes. But then, I noticed one property of initial topology as follows:
If $X$ has the initial topology generated by a family $\mathcal{F}$ of functions,then a net $\langle x_\alpha\rangle$ converges to $x \in X$ iff $\langle f(x_\alpha)\rangle$ converges to $f(x)$ for all $f \in \mathcal{F}$.
This would imply that a sequence $\{ x_n :n \in \mathbb{N}\}$ converges to $x \in X$ iff $\| x_n\|$ converges to $\|x\|$, which in general I think is not true.
Why am I getting this contradiction? Where did I go wrong? 
Thanks in advance for any help!
 A: What you have found is that the norm topology is not the initial topology with respect to the norm function.
Another way to see that the usual norm topology is not the initial topology is as follows: let $n:X \to \Bbb R$ denote the norm function $n(x) = \|x\|$.  Fix any $x_1,x_2$ such that $\|x_1\| = \|x_2\| = 1$.  Then for any $U = (a,b) \subset \Bbb R$, we have $x_1 \in n^{-1}(U) \iff x_2 \in n^{-1}(U)$.  Thus, in the initial topology, every neighborhood of $x_1$ is a neighborhood of $x_2$ and vice versa, which is to say that the points are topologically indistinguishable.  Your initial topology is not even Hausdorff.
It is correct, however, to say that the usual norm topology is the initial topology with respect to the family of functions $\mathcal F = \{n_x : x \in X\}$ where
$$
n_x(y) = \|x - y\|
$$
Alternatively, perhaps we may describe it as the coarsest topology such that $n(x)$ and the map $x,y \mapsto x-y$ (from $X \times X$ to $X$) is continuous.
A: The topology on $X$ is not generated by the norm function in this manner.
The open sets generated by $\| \cdot \|_X : X \to \mathbb{R}$ are just the open balls and open 'shells' centred at the origin. This does not even define a Hausdorff topology. Indeed if $\|x\| = \|y\|$ then an open set contains $x$ if and only if it contains $y$.
What is true is that the topology on a metric space $(M,d)$ is the weakest topology such that the map $d \colon M \times M \to \mathbb R$ is continuous. 
