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Let $E$ be a $\mathbb R$-vector space, $P$ be a nonempty family of seminorms on $E$, $$U_p:=\{x\in E:p(x)<1\}\;\;\;\text{for }p\in P$$ and $\tau$ denote the topology on $E$ generated by $P$.

I was able to show that \begin{equation}\begin{split}\mathcal B_P&:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}\\&=\left\{\bigcap_{p\in F}U_{\frac1\varepsilon p}:F\subseteq P\text{ is finite and }\varepsilon>0\right\}\end{split}\end{equation} is an analytic basis$^1$ for $\tau$.

Now I was wondering whether $$\mathcal S_P:=\left\{\varepsilon U_p:p\in P\text{ and }\varepsilon>0\right\}=\left\{U_{\frac1\varepsilon p}:p\in P\text{ and }\varepsilon>0\right\}$$ is an analytic subbasis$^2$ for $\tau$.

By definition, this would imply that $$\mathcal B'_P:=\left\{\bigcap\mathcal F:\mathcal F\subseteq\mathcal S_P\text{ is finite}\right\}\tag1$$ is an analytic basis for $\tau$.

Does this in turn imply that $\mathcal B_P=\mathcal B'_P$?

The inclusion $\mathcal B_P\subseteq\mathcal B'_P$ is clearly trivial. Let's consider the other inclusion: Let $B\in\mathcal B_P'$. Then there is a finite $\mathcal F\subseteq\mathcal S_P$ with $$B=\bigcap\mathcal F.$$ Assume $n:=|\mathcal F|\in\mathbb N$ (i.e. $\mathcal F\ne\emptyset$). Since $\mathcal F\subseteq\mathcal S_P$, $$\mathcal F=\left\{\varepsilon_1U_{p_1},\ldots,\varepsilon_nU_{p_n}\right\}$$ for some $p_1,\ldots,p_n\in P$ and $\varepsilon_1,\ldots,\varepsilon_n>0$. If $$p:=\max_{1\le i\le n}\frac1{\varepsilon_i}p_i,$$ then clearly $$B=\bigcap_{i=1}^n\varepsilon_iU_{p_i}=U_p\tag2.$$

However, that doesn't immediately yield $B\in\mathcal B_P$, unless $p\in P$. So, maybe the desired claim doesn't hold and $\mathcal B_P$ and $\mathcal B'_p$ are distinct analytic bases for $\tau$. If that's the case, is my initial guess that $\mathcal S_P$ is in fact an analytic subbasis for $\tau$ correct at all?

$^1$ $\mathcal B$ is called analytic basis for $\tau$ if $\mathcal B\subseteq2^E$ and $$\tau=\left\{\bigcup\mathcal A:\mathcal A\subseteq\mathcal B\right\}.$$

$^2$ $\mathcal S$ is called analytic subbasis for $\tau$ if $\mathcal S\subseteq2^E$ and $$\left\{\bigcap\mathcal F:\mathcal F\subseteq\mathcal S\text{ is finite}\right\}$$ is an analytic basis for $\tau$.

mechanodroid
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0xbadf00d
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1 Answers1

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As you noticed, it doesn't have to be $\mathcal{B}_P = \mathcal{B}_P'$ in general. However, $\mathcal{B}_P'$ is an analytic basis for $\tau$. Indeed, clearly we have $$\mathcal{B}_P \subseteq \mathcal{B}_P' \subseteq \tau$$ because $\varepsilon U_p$ are in $\tau$ and $\tau$ is closed under finite intersections. In particular

  • for any $\mathcal{A} \subseteq \mathcal{B}_P'$ we have $\bigcup \mathcal{A} \in \tau$ since $\mathcal{B}_P' \subseteq \tau$ and $\tau$ is closed under unions.
  • for any $U \in \tau$ since $\mathcal{B}_P$ is an analytic basis there exists $\mathcal{A} \subseteq \mathcal{B}_P \subseteq \mathcal{B}_P'$ such that $U =\bigcup\mathcal{A}$.
mechanodroid
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  • Thank you for your answer. It remains to show that $\mathcal S_P$ is actually an analytic subbasis for $\tau$. How do we see that? – 0xbadf00d Dec 18 '20 at 10:25
  • @0xbadf00d By your definition, it is sufficient to show that $\mathcal{B}_P'$ is an analytic basis for $\tau$, which is exactly what we showed. – mechanodroid Dec 18 '20 at 10:35
  • Yes, sure. I thought there might be a nice direct argument which doesn't require to know that $\mathcal B_P$ is an analytic basis. – 0xbadf00d Dec 18 '20 at 10:45
  • @0xbadf00d Maybe there is, how exactly do you define $\tau$ from $P$? – mechanodroid Dec 18 '20 at 10:51
  • In general, if $E$ is a set, $(E',\tau')$ is a topological space and $f:E\to E'$, the topology generated by $f$ is defined to be $$\tau(f):={f^{-1}(\Omega'):\Omega'\in\tau'}.$$ And if $I$ is a nonempty set and $E_i$ is a topological space and $f_i:E\to E_i$ for $i\in I$, the topology generated by $(f_i)_{i\in I}$ is defined to be $$\tau(f_i,i\in I):=\tau\left(\bigcup_{i\in I}\tau(f_i)\right).$$ With this definition, in the situation of the question, $$\tau:=\tau(P).$$ – 0xbadf00d Dec 18 '20 at 11:03
  • Well, you can simply reproduce how you proved that $\mathcal{B}_P$ is an analytic basis since $\mathcal{B}_P'$ contains $\mathcal{B}_P$. – mechanodroid Dec 18 '20 at 11:17
  • I didn't thought on the most simple case where $P$ consists of a single norm before, but now I wonder why $\mathcal B_P$ is actually an analytic basis. Please take a look at this question: https://math.stackexchange.com/q/3938301/47771. – 0xbadf00d Jan 04 '21 at 08:28