How to show that the topology is compatible with the metric?

Question

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!

Answer 1

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:

1. 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 $B_R(y)\subset U$.

2. 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:

2. 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.

Answer 2

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?