To your second question: The defined topology is the smallest topological vector space (TVS) topology so that all semi-norms are continuous.
To understand why the neighborhood bases are defined this way (and why such a definition is natural) a review of the topology of TVS's is in order:
We need the following proposition (for a proof see here):
Proposition: Let $X$ be a set and $\mathcal{N} : X \to \mathcal{P}(\mathcal{P}(X))$ have the following properties for all $x \in X$:
- $\mathcal{N}(x) $ is a filter
- $ x \in N$ for all $N \in \mathcal{N}(x)$
- The set $\{ z \in N : N \in \mathcal{N}(z) \} $ is in $\mathcal{N}(x)$ for all $N \in \mathcal{N}(x)$
Then the set
$$\tau = \{ U \subset X: U \in \mathcal{N}(x) \, \forall x \in U \}$$
defines a topology on $X$ and the below defined neighborhood system agrees with $\mathcal{N}$. (So the bizzare looking third condition actually says that the interior of every neighborhood $N$ of a point $x$ is a neighborhoood itself.)
Conversely if $\tau$ is a topology on $X$ then the neighborhood system $\mathcal{N}: X \to \mathcal{P}(\mathcal{P}(X))$ defined by
$$\mathcal{N}(x)= \{ N \subset X : \exists U \in \tau \ \text{with} \ x \in U\subset N \}$$
satisfies the above properties and the topology induced by it is the same as the original topology.
For a TVS $X$ it is true that $\mathcal{N}(x) = x + \mathcal{N}(0)$ (here $\mathcal{N}$ is the neighborhood system of the TVS) for all $x \in X$, because $+$ is a homeomorphism. Therefore the topology is fully determined by the neighborhood filter of $0$.
Define a filter base (of the neighborhood filter at $0$) to be a subset $\mathcal{B} \subset \mathcal{N}(0)$ so that
$$\mathcal{N}(0) = \{ N \subset X : \exists B \in \mathcal{B} \ \text{with} \ B \subset N \}.$$
A filter base (of the neighborhood filter at $0$) must satisfies the following:
- $\mathcal{B} \neq \varnothing$
- $A \in \mathcal{B} \Rightarrow 0 \in A$
- $A, B \in \mathcal{B} \Rightarrow \exists C \in \mathcal{B} : C \subset A \cap B$
and conversely any $\mathcal{B} \subset\mathcal{P}(X)$ with the above properties is a filter base for the filter of sets containing $0$ defined by
$$\mathcal{N}(0) = \{ N \subset X : \exists B \in \mathcal{B} \ \text{with} \ B \subset N \}.$$
The next theorem tells us that for a TVS we only need to specify a filter base with certain properties:
Theorem:
Let $X$ be a vector space. Let $\mathcal{N}(0)$ be a filter of sets containing $0$ that has a filter base $\mathcal{B}$ (as defined above) with the following properties:
- $B \in \mathcal{B} \Rightarrow \exists A\in \mathcal{B} : A +A \subset B $
- every subset of $ \mathcal{B}$ is balanced and absorbing
Then the topology induced (as in the above proposition) by the system $x \mapsto \mathcal{N}(x)= x+ \mathcal{N}(0)$ is a TVS topology and $\mathcal{N}(0)$ is the neighborhood filter of $0$. Conversely given a TVS topology on $X$ the neighborhood filter satisfies the above (and induces the same topology).
For a statement of this theorem with source see here.
Furthermore a semi-norm is continuous with respect to a TVS topology if and only if it is continuous at $0$ (follows from the reverse triangle inequality).
Therefore it is sufficient and necessary that $B_\alpha(r) = p_\alpha^{-1}([0,r)) \in \mathcal{N}(0)$ for any $\alpha \in I ,r >0$ for the continuity of the semi-norms.
Lets consider the case of a single semi-norm $p$ first. Then it is not hard to see that
$$ \mathcal{B}= \{ B(r) : r > 0 \}$$
is a filter base that satisfies all the assumptions of the theorem. The induced topology is also the smallest TVS topology for which the semi-norm is continuous. Since for a second TVS topology in which the semi-norm is continuous there is for every $B \in \mathcal{B}$ an open (in the second topology) set $U$ with $0 \in U \subset B$ and so if a set is a neighborhood of all of its points in the first topology then it also is in the second.
Now if there are 2 semi-norms $p_1, p_2$, then maybe we want to define
$$ \mathcal{B}= \{ B_{\alpha}(r) : r > 0, \alpha = 1,2 \}.$$
But this is not a filter base, because given $B_1 (r_1) $ and $B_2(r_2)$ there is in general no ball $B \in \mathcal{B}$ with $B \subset B_1 (r_1) \cap B_2(r_2)$.
Now the obvious fix is to simply include these intersections:
$$ \mathcal{B}= \{ \cap_{j=1}^n B_{\alpha_j}(r_j) : n \in \{ 1,2 \}, r_j > 0, \alpha_j \in \{1,2 \} \}.$$
It is not hard to check either that $\mathcal{B}$ satisfies all the conditions in the theorem and that this yields the minimal topology (since the finite intersection of open sets containing $0$ is an open set containing $0$). This should answer your first question.
The same idea applies to any number of semi-norms and we arrive at the filter base described in your post.