A question about a family of seminorms inducing topology

Question

I have a question from Rudin's Functional Analysis. The theorem's statement and the proof go like the following.

In the proof, it says "a set $A \subset X$ to be open if and only if $A$ is a union of translates of members of $\mathscr{B}$". However, for the simplest case, given two open sets $\big( x+V(p,1/n) \big)$and $\big(y+V(q,1/m)\big)$, their intersection is open?

In other words, can we find another translate of a member of $\mathscr{B}$ which contained in $\big(x+V(p,1/n)\big) \cap \big(y+V(q,1/m)\big)$? I think it is equivalent to the statement that the translates of every member of $\mathscr{B}$ are the base of a topology.

I'd appreciate any help! Thank you.

Answer 1

Suppose that $(x+V(p,1/n))\cap(y+V(q,1/m))$ is nonempty and let $$ z\in(x+V(p,1/n))\cap(y+V(q,1/m)), $$ that is, $$ p(z-x)<1/n,\qquad q(z-y)<1/m. $$ There exists $\varepsilon>0$ such that $$ p(z-x)+\varepsilon<1/n,\qquad q(z-y)+\varepsilon<1/m. $$ Hence, if $\xi\in z+\{p<\varepsilon\}\cap\{q<\varepsilon\}$ then $$ p(\xi-x)<p(\xi-z)+p(z-x)<\varepsilon+p(z-x)<1/n, $$ and $$ q(\xi-y)<q(\xi-z)+q(z-y)<\varepsilon+q(z-y)<1/m, $$ meaning that $$ \xi\in(x+V(p,1/n))\cap(y+V(q,1/m)). $$

Note: Another answer appeared while I was writing this answer. I leave it here since it has a tiny bit more details.

Answer 2

Let $z\in U:=(x+V(p,n))\cap (y+V(q,m))$ and choose $N> (\frac{1}{n}-p(z-x))^{-1}+(\frac{1}{m}-q(z-y))^{-1}$. Then, for any $w\in z+V(p,N)\cap V(q,N)$, we have $p(w-x)\leq p(w-z)+p(z-x)<\frac{1}{n}$. Similarly, we have $q(w-y)<\frac{1}{m}$. This establishes that $z+V(p,N)\cap V(q,n)\subseteq U$. Since $z$ was arbitrary, we get that $U$ is, indeed, open in the proposed topology.