How to show that the topology is compatible with the metric? This is in the contest of Toplogical-Vector-Spaces, but can be interepreted as a simply topology question.
For my matter, assume $\|\cdot\|_n$ is a countable family of seminorms, and define $$d(x,y)=\sum 2^{-n} \frac{\|x-y\|_n}{1+\|x-y\|_n}.$$
It is quite easy to see that this is a metric, and I want to show that it is compatible with the topology generated by this local-sub-base at $0$:
$$\left\{x\in X\mid\:\|x\|_i < \frac{1}{n}\right\}\text{ (for all }i,n\text{)}$$
How am I supposed to do this? By definition, I believe i'm supposed to show that they have the same open sets, but pointing out some kind of function between those doesn't seem right to me.
Thanks!
 A: All you need to do is prove that the two topologies (the metric topology, and the one generated by the sub-base) are equivalent, which is to prove the following:


*

*Let $U$ be open in the topology generated by the local sub-base. Then there is an open ball $B_R(y) = \{x\in X: d(x,y)<R\}$ with $B_R(y)\subset U$.

*Let $B_R(y)$ be an open ball as defined above. Show that there is an open set $U$ in the sub-base topology with $U\subset B_R(y)$.
As for how to do this:
You can do 1. on the sub-base level, i.e. let $U$ be in the sub-base. It shouldn't be all that hard to guess the appropriate $R$. 2. is the harder direction.
Remark:
This isn't a question you should think of as purely topological. The vector space structure makes things easier, namely you should be able to do everything at $0$. It's important, but not all that hard, to go through the details of translating the argument to the rest of the space.
Edit:


*is not so hard if you know that you can convert a family of seminorms into a directed (in the notation of Reed & Simon) family of seminorms generating an equivalent topology.

A: Fix $ \epsilon > 0 $, and let $ \mathbb{B}(0;\epsilon) $ denote the open $ d $-ball of radius $ \epsilon $ whose center is $ 0 $. Choose $ x^{*} \in \mathbb{B}(0;\epsilon) $. Then $ \delta \stackrel{\text{df}}{=} d(0,x^{*}) < \epsilon $.
Next, pick $ N \in \mathbb{N} $ large enough so that
$$
\sum_{n = N + 1}^{\infty} 2^{- n} < \frac{\epsilon - \delta}{2}.
$$
Let $ U $ denote the following subset of $ X $ that is open with respect to the semi-norm topology:
$$
\left\{
x \in X
~ \middle| ~
\forall n \in [N]: \quad \| x - x^{*} \|_{n} < \dfrac{\epsilon - \delta}{2 N}
\right\}.
$$
Then for all $ x \in U $, we have
\begin{align}
    d(x,x^{*})
& = \sum_{n = 1}^{\infty}
    2^{- n} \cdot \frac{\| x - x^{*} \|_{n}}{1 + \| x - x^{*} \|_{n}} \\
& = \sum_{n = 1}^{N} 2^{- n} \cdot \frac{\| x - x^{*} \|_{n}}{1 + \| x - x^{*} \|_{n}} +
    \sum_{n = N + 1}^{\infty}
    2^{- n} \cdot \frac{\| x - x^{*} \|_{n}}{1 + \| x - x^{*} \|_{n}} \\
& < \sum_{n = 1}^{N} \| x - x^{*} \|_{n} + \sum_{n = N + 1}^{\infty} 2^{- n} \\
& < \sum_{n = 1}^{N} \frac{\epsilon - \delta}{2 N} + \frac{\epsilon - \delta}{2} \\
& = \epsilon - \delta,
\end{align}
which means that
$$
d(0,x) \leq d(0,x^{*}) + d(x^{*},x) < \delta + (\epsilon - \delta) = \epsilon.
$$
Hence, $ x^{*} \in U \subseteq \mathbb{B}(0;\epsilon) $, and it follows that $ \mathbb{B}(0;\epsilon) $ is a union of sets that are open with respect to the semi-norm topology. Therefore, $ \mathbb{B}(0;\epsilon) $ itself is open with respect to the semi-norm topology.
You can prove the converse via a similar $ \epsilon $-argument. Would you like to give it a try first?
