Subbase for initial topology of quotient mappings (locally convex spaces)

Question

Consider a family $\{p_i\}_{i\in I}$ of seminorms defined on $X$ and consider for each $i\in I$ the quotient mapping $q_i:X\rightarrow X/\ker(p_i)$. I know, that each quotient $X/\ker(p_i)$ is a vector space (turning the quotient map into a linear map) and has a norm, defined by $$\lVert q_i(x)\rVert:=\inf_{y\in \ker(p_i)}p_i(x+y)$$

Now, the author states:

For $i\in I$, $\epsilon> 0$ and $x\in E$, the sets $$B_{\epsilon}^i(x)=\{ z\in X : p_i(x-z)<\epsilon \}$$ form a subbase of the initial topology of the family $\{q_i:X\rightarrow X/\ker(p_i)\}_{i\in I}$, where each quotient carries the normed topology.

My attempt is the following:

The normed topology on $X/\ker(p_i)$ has a base consisting of sets

$$B_\epsilon(q_i(x))=\{z\in X/\ker(p_i):\lVert q_i(x)-z\rVert<\epsilon\}$$

and it's preimage (by linearity of $q_i$) is

$$q_i^{-1}(B_\epsilon(q_i(x)))=\{z\in X:\lVert q_i(x-z)\rVert<\epsilon\}.$$

Now, it holds $B^i_\epsilon(x)\subseteq q_i^{-1}(B_\epsilon(q_i(x)))$, since $0\in\ker(p_i)$ and therefore

$$\epsilon > p_i(x-z)=p_i((x-z)+0)\ge\inf_{y\in\ker(p_i)}p_i((x-z)+y)=\lVert q_i(x-z)\rVert.$$

Unfortunately, I am stuck here. Can I possibly even show equality? Or can I describe $q_i^{-1}(B_\epsilon(q_i(x)))$ as union of these $B^i_\epsilon(y)$'s?

Any hint or help is appreciated! Thank you in advance!

EDIT: For $B^i_\epsilon(x)\supseteq q_i^{-1}(B_\epsilon(q_i(x)))$, one can observe that

$$\epsilon>\lVert q_i(x-z)\rVert\ge p_i(x-z).$$

Suppose the contrary, i.e. $\inf_{y\in\ker(p_i)}p_i((x-z)+y)=\lVert q_i(x-z)\rVert< p_i(x-z)$. Then there exists a $y\in\ker(p_i)$, such that $p_i((x-z)+y)< p_i(x-z)$ and therefore $$ \begin{aligned} 0<p_i(x-z)-p_i((x-z)+y) &=|p_i(x-z)-p_i((x-z)+y)|\\ &\le|p_i(x-z-(x-z)-y)|\\ &=|p_i(y)|=0, \end{aligned}$$ by using the reversed triangle inequality and the fact that $y\in \ker(p_i)$. This is contradictory and therefore it holds $$\lVert q_i(x-z)\rVert\ge p_i(x-z).$$

In fact, from the calculations above, one can also just observe, that

$$\lVert q_i\rVert=p_i,$$ such that

$$B_\epsilon^i(x)=\{z\in X: p_i(x-z)<\epsilon\}=\{z\in X: \lVert q_i(x-z)\rVert<\epsilon\}=q_i^{-1}(B_\epsilon(q_i(x))).$$

Since the $\epsilon$-balls are a base of $\mathbb{R}_{\ge 0}$, the collection of its preimages is even a base (and therefore in fact a subbase).

Answer 1

Fix $y=q_i(x)\in X_i=X/\ker p_i$ and $\epsilon>0$. We first want to show that $U:=q_i^{-1}(B_\epsilon(y))$ is open in the topology generated by the $(B^i_\epsilon(x))_{i\in I,\epsilon>0,x\in X}$.

Take $a\in U$. I know $\inf_{b\in\ker p_i}p_i(a+b-x)<\epsilon$ so in particular there must be at least one $b\in\ker p_i$ with $\delta:=p_i(a+b-x)<\epsilon$. Define $\eta:=\frac{\epsilon-\delta}{2}>0$. I claim $B^i_\eta(a)\subseteq U$: take $a'\in B^i_\eta(a)$.

We are told $p_i(a'-a)<\eta$, and we want to show $p_i(a'+b'-x)<\epsilon$ for some $b'\in\ker p_i$. Well, I can choose $b'=b$ and have $p_i(a'+b-x)\le p_i(a'-a)+p_i(a+b-x)<\epsilon$, so the claim holds. Great! It follows that $U$ is open since $a$ was arbitrary.

Now we want to show that every single $B^i_\epsilon$ is open in the topology generated by the sets of the form of $U$ (the initial topology). Fix $x\in X$ and $i,\epsilon$ and some $x'\in B^i_\epsilon(x)$. Define $\delta:=p_i(x-x')<\epsilon$ and $\eta:=\frac{\epsilon-\delta}{2}>0$.

Take $z\in q_i^{-1}(B_\eta(q_i(x')))$. That means $p_i(z+b-x')<\epsilon$ for some $b\in\ker p_i$ and it follows that $p_i(z-x)\le p_i(z+b-x')+p_i(x'-x-b)\le p_i(z+b-x')+p_i(x'-x)<\epsilon$. Hence $q_i^{-1}(B_\eta(q_i(x')))\subseteq B^i_\epsilon(x)$ and because $x'$ was arbitrary we conclude $B^i_\epsilon(x)$ is open in the initial topology.

Therefore the two topologies coincide.