I have the following task. Let $X$ be a vector space equipped with a family $\mathcal{P} = \{\rho_{n}\}_{n\in \mathbb{N}}$ of semi-norms. I want to prove that:
- This family of semi-norms generate a topology on $X$ and
- this topology generated by $\mathcal{P}$ is the same as the one generated by the metric: $$d(x,y) = \sum_{n=1}^{\infty}2^{-n}\frac{\rho_{n}(x-y)}{1+\rho_{n}(x-y)}$$
My attempt to 1. According to Engelking's General Topology book, we have the following proposition.
Proposition: Suppose we are given a set $X$ and a collection $\{\mathcal{B}(x)\}_{x\in X}$ of families of subsets of $X$ satisfying the following properties:
(BP1) For every $x \in X$, $\mathcal{B}(x) \neq \emptyset$ and for every $U \in \mathcal{B}(x)$, $x \in U$.
(BP2) If $x \in U \in \mathcal{B}(y)$, then there exists a $V \in \mathcal{B}(x)$ such that $V \subset U$
(BP3) For any $U_{1}, U_{2} \in \mathcal{B}(x)$ there exists a $U \in \mathcal{B}(x)$ such that $U \subset U_{1}\cap U_{2}$. Then, the family $\mathcal{O} = \bigcup_{x\in X}\mathcal{B}(x)$ is a topology on $X$ and the collection $\{\mathcal{B}(x)\}_{x\in X}$ is a neighborhood system for this topological space.
Given $x \in X$, $n_{1},...,n_{k} \in \mathbb{N}$ and $\epsilon > 0$, consider the set: $$U_{n_{1},...,n_{k}}(x;\epsilon) := \{y \in X: \text{$\rho_{n_{i}}(y-x) < \epsilon$ for every $i=1,...,k$}\}$$ Let $\mathcal{B}(x)$ be the set of all such sets. I have to prove that the latter satisfies properties (BP1) to (BP3). Since $\rho_{n}(0) = 0$ for every $n \in \mathbb{N}$, (BP1) is easily verified. Suppose $x \in U = U_{n_{1},...,n_{k}}(y;\epsilon) \in \mathcal{B}(y)$. Take $\tilde{\epsilon} = \epsilon - \max_{i \in \{1,...,k\}}\rho_{n_{i}}(x-y)$ and consider $U_{n_{1},...,n_{k}}(x;\tilde{\epsilon})$. Then, if $z \in U_{n_{1},...,n_{k}}(x,\tilde{\epsilon})$ we have: $$\rho_{n_{i}}(z-y) \le \rho_{n_{i}}(z-x) + \rho_{n_{i}}(x-y) < \epsilon -\max_{i}\rho_{n_{i}}(x-y)+\max_{i}\rho_{n_{i}}(x-y) < \epsilon$$ This proves (BP2). Finally, if $U = U_{n_{1},...,n_{k}}(x;\epsilon), V = V_{m_{1},...,m_{l}}(x;\delta) \in \mathcal{B}(x)$, then $U_{n_{1},...,n_{k},m_{1},...,m_{l}}(x,\min\{\epsilon,\delta\}) \subset U\cap V$, which proves (BP3).
Attempt to item 2. Suppose $x \in U_{n_{1},...,n_{k}}(x;\epsilon)$ and take $V_{\epsilon} = \{y \in X: d(x,y) < \epsilon\}$. Then: $$y \in V_{\epsilon} \iff \sum_{n=1}^{\infty}2^{-n}\frac{\rho_{n}(x-y)}{1+\rho_{n}(x-y)}< \epsilon \Rightarrow \rho_{n}(x-y) < \epsilon $$ and $V_{\epsilon} \subset U_{n_{1},...,n_{k}}(x;\epsilon)$.
Conversely, let $V_{\epsilon}$ be given as before. Given $\delta > 0$, there exists $N_{0} \in \mathbb{N}$ such that: $$\sum_{n=N}^{\infty}2^{-n} < \delta$$ for every $N \ge N_{0}$. Take $\tilde{\epsilon} = \frac{2^{N}}{N}(\epsilon - \delta)$ and $U = \{y \in X: \text{$\rho_{n}(y-x) < \tilde{\epsilon}% for every $n=1,...,N$\}$. Then if $y \in U$ we have: $$\sum_{n=1}^{\infty}2^{-n}\frac{\rho_{n}(x-y)}{1+\rho_{n}(x-y)} = \sum_{n=1}^{N}2^{-n}\frac{\rho_{n}(x-y)}{1+\rho_{n}(x-y)} + \sum_{n=N}^{\infty}2^{-n}\frac{\rho_{n}(x-y)}{1+\rho_{n}(x-y)} \le \sum_{n=1}^{N}2^{-n}\frac{2^{N}}{N}(\epsilon - \delta) + \sum_{n=N}^{\infty}2^{-n} < \epsilon - \delta + \delta = \epsilon$$ so that $U\subset V_{\epsilon}$.
Question: Are my proofs correct? I am not entirely sure about my reasoning the second part of item 2. and also it is my first time proving something in the lines of part 1. so I would like to know if I did everything right.