Please help me escape from my tantalizing confusion. In the proof, it is claimed that $\mathcal{B}$ induces an translation-invariant topology $\tau$.
The translation-invariance of the collection $\tau$ is obvious by construction. And, in order to prove that the collection $\tau$ is a topology, it would suffice to show that $A\cap B$ is open whenever $A$ and $B$ are both open. (The other requirements for $\tau$ to be a topology seem trivial.)
But I was stuck with this, and thus referring to many other online lecture notes, I received the impression that there may be another logic for showing that $\tau$ is a topology, because no single one tried in my unavailing way.
In this sense, is it enough to show that, for any $A \in \tau$ containing $0_X$, there exists some $\beta \in \mathcal{B}$ such that $\beta \subset A$? But this feels awkward because we do not yet know that $\tau$ is a topology. Most lecture notes seem to follow this logic, which I cannot agree with easily.
I know that, for any topological vector space $(X, \tau)$, the vector topology $\tau$ is fully characterized by any local base $\mathcal{B}$, due to translation-invariance. But in this case we already know that $\tau$ is a topology! My current understanding for the most lecture notes is that they are claiming that a collection $\tau$ of subsets of $X$ is turned by $\mathcal{B}$ into a topology on $X$ if only $\mathcal{B}$ plays a role of a local base for $\tau$ as mentioned above.
Any clarifying comment would be enormously appreciated!
I propose a slightly alternative but equivalent way of defining $\tau$:
$$U \in \tau \iff \forall x \in U: \exists B_x \in \mathcal{B}: x+B_x \subseteq U\tag{1}$$
This is similar how open sets are defined in metric spaces using balls. It's IMHO easier to check this definition yields a topology, I'll do intersections of two open sets as an example:
Suppose $U,V \in \tau$. Suppose that $x \in U \cap V$, so $x + B_1\subseteq U$ and $x + B_2 \subseteq V$ where $B_1,B_2 \in \mathcal{B}$, and then note that by definition $\mathcal{B}$ is closed under finite intersections (as it is itself defined as a set of finite intersections of sets of the form $V(p,n)$, where $p \in \mathcal{P}$ and $n \in \mathbb{N}$) so $B_1 \cap B_2 \in \mathcal{B}$ and clearly $x + (B_1 \cap B_2) \subseteq U \cap V$. (This is analogous to taking the minimum of two radii in the metric context, if you think about it).
Closedness under all unions and $\emptyset,X \in \tau$ are trivial. It's also clear that "my" $\tau$ definition also defines a translation invariant topology with local base $\mathcal{B}$, so it's the same one that Rudin defined in his own way (same local base at $0$!).