TITLE: Topology induced by seminorms and initial topology
QUESTION [8 upvotes]: Let's say we have a family of seminorms $(\rho_\alpha)_{\alpha \in A}$ on
a vector space $V$.
There are two ways to topologize $V$ using those seminorms:

We define topology $\mathcal S$ by a subbasis consisting of sets of
the form
$$ B_{x, \alpha, \epsilon} =\{v \in V : \rho_\alpha(v - x) < \epsilon \} $$
for some $\epsilon > 0, x \in V, \alpha \in A$.
We define topology $\mathcal I$ as the smallest topology in which
the seminorms are continuous.

I've showed that the seminorms are continuous in $\mathcal S$, and that 
implies $\mathcal I \subseteq \mathcal S$.
I'm having trouble with the other direction.
It is clear to me that $B_{0, \alpha, \epsilon} = \rho_\alpha^{-1} 
((-\epsilon, \epsilon ))$, so it's enough to show that vector addition
is continuous in respect to $\mathcal I$, but I don't know how to do that.

REPLY [9 votes]: Apart from trivial cases - all seminorms are identically $0$ - the two topologies don't coincide.
A seminorm cannot distinguish $x$ and $-x$, so every neighbourhood of $x$ in $\mathcal{I}$ contains $-x$. For an $x$ and a seminorm $p$ with $p(x) \neq 0$, there is however a ball $\{ x : p(y-x) < \epsilon\}$ that doesn't contain $-x$.
If you take the initial topology with respect to the family
$$\{ x \mapsto \rho_\alpha(x-y) : \alpha \in A, y \in V\},$$
then you describe $\mathcal{S}$ as an initial topology.