0

Let $(X,\tau)$ be a topological space, $Y$ be a normed vector space $$\overline p(f):=1\wedge\sup_{x\in X}\left\|f(x)\right\|_Y\;\;\;\text{for }f\in C(X,\tau;Y)$$ and $$p_K(f):=\sup_{x\in K}\left\|f(x)\right\|_Y\;\;\;\text{for }C(X,\tau;Y)$$ for $\tau$-compact $K\subseteq X$ and $$P:=\left\{p_K:K\text{ is }\tau\text{-compact}\right\}.$$ We know (see below) that $$\mathcal B_{\overline p}:=\left\{\varepsilon U_{\overline p}:\varepsilon>0\right\}$$ is a local basis for the topology $\tau(\overline p)$ generated by $\overline p$ on $C(X,\tau;Y)$ at $0$ and $$\mathcal B_P:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}$$ is an analytic basis for the topology $\tau(P)$ generated by $P$ on $C(X,\tau;Y)$ at $0$.

It's trivial to see that $$U_{\overline p}=\left\{f\in C(X,\tau;Y):\sup_{x\in X}\left\|f(x)\right\|_Y<1\right\}\subseteq\bigcap_{p\in P}U_p\tag1.$$ So, $$\mathcal B_{\overline p}\subseteq\mathcal B_P\tag2.$$

There must be something crucial that I'm missing, since $(2)$ should immediately imply that $\tau(\overline p)$ is coarser than $\tau(P)$, i.e. $\tau(\overline p)\subseteq\tau(P)$. The same argumentation would yield that the uniform operator topology is coarser than the strong operator topology; but we know that it's the other way around. So, what am I missing?

Let $(E,\tau)$ be a topological space.

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

  2. Let $x\in E$ and $$\mathcal N_\tau(x):=\left\{N\subseteq E:N\text{ is a }\tau\text{-neighbhorhood of }x\right\}.$$ $\mathcal B$ is called neighbhorhood basis for $\tau$ at $x$ if $\mathcal B\subseteq\mathcal N_\tau(x)$ and $$\mathcal N_\tau(x)=\left\{N\subseteq E:\exists B\in\mathcal B:B\subseteq N\right\}.$$ In that case, it is called local basis for $\tau$ at $x$ if $\mathcal B\subseteq\tau$. $\mathcal S$ is called neighbhorhood subbasis for $\tau$ at $x$ if $\mathcal S\subseteq\mathcal N_\tau(x)$ and $$\left\{\bigcap\mathcal F:\mathcal F\text{ is finite}\right\}$$ is a neighborhood basis for $\tau$ at $x$. In that case, it is called local subbasis for $\tau$ at $x$ if $\mathcal S\subseteq\tau$.

Assume $E$ is a vector space and $\tau$ is a vector topology on $E$.

  1. If $x\in E$, then $\mathcal N_\tau(x)=x+\mathcal N_\tau(0)$ and if $\mathcal B$ is a neighborhood basis for $\tau$ at $0$, then $x+B$ is a neighborhood basis for $\tau$ at $x$.

  2. If $\mathcal B(x)$ is a neighborhood basis for $\tau$ at $x$ for $x\in E$ and $$\mathcal B:=\bigcup_{x\in E}\mathcal B(x),$$ then $$\tau\subseteq\left\{\bigcup\mathcal A:\mathcal A\subseteq\mathcal B\right\}.$$ If $\mathcal B(x)$ is a local basis for $\tau$ at $x$ for all $x\in E$, then $\mathcal B$ is an analytic basis for $\tau$.

Assume $\tau$ is generated by a family $P$ of seminorms on $E$. Let $U_p:=\{x\in E:p(x)<1\}$ for $p\in P$.

  1. $$\mathcal B_P:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}$$ is a local basis for $\tau$ at $0$. $$\mathcal S_P:=\left\{\varepsilon U_p:p\in P\text{ and }\varepsilon>0\right\}$$ is a local subbasis for $\tau$ at $0$.
0xbadf00d
  • 14,208
  • 1
    You have forgotten to translate your sets. Your system only generates the neighbourhoods of $0$. – s.harp Jan 04 '21 at 10:21
  • @s.harp Thanks for mentioning. You're right, I've confused the concepts with the notions of a neighborhood/local basis. I've completely rewritten the question. There must still be something I'm missing, since for the moment my understanding would yield that the uniform operator topology is coarser than the strong operator topology; but I know that it's the other way around. – 0xbadf00d Jan 04 '21 at 11:52
  • 1
    If you have $\mathcal{B_1,B_2}$ two local basis of $0$ and for every $U_1\in \mathcal B_1$ that there is a $U_2\in\mathcal B_2$ with $U_1\subseteq U_2$ then the neighbourhoods given by $\mathcal B_1$ are finer than $\mathcal B_2$, not coarser. The mistake you have made is assuming that $U_1\subseteq U_2$ implies $\mathcal B_1\subseteq \mathcal B_2$. Additionally you may have thought $$U_{\overline p}=\bigcap_{p\in P}U_p$$ to directly imply $\mathcal B_1 \subseteq \mathcal B_2$, but note that this intersection is not a finite intersection! – s.harp Jan 04 '21 at 12:07
  • @s.harp I guess you're right. If $B\in\mathcal B_{\overline p}$, then $B=\varepsilon U_{\overline p}$ for some $\varepsilon>0$. If $\varepsilon>1$, then $B=C(X,\tau;Y)\in\mathcal B_P$. If $\varepsilon\le1$, we can take an arbitrary $\tau$-compact $K\subseteq X$ and notice that $$B=\left{f\in C(X,\tau;Y):\sup_{x\in X}\left|f(x)\right|Y<\varepsilon\right}\subseteq\varepsilon U{p_K}\in\mathcal B_P.$$ – 0xbadf00d Jan 04 '21 at 18:28
  • @s.harp So, we have shown $$\forall B_1\in\mathcal B_{\overline p}:\exists B_2\in\mathcal B_P:B_1\subseteq B_2.\tag3$$ Now $O\subseteq C(X,\tau;Y)$ is $\tau(\overline p)$-open iff $$\forall x\in O:\exists B_1\in\mathcal B_{\overline p}:x+B_1\subseteq O\tag4$$ and $\tau(P)$-open iff $$\forall x\in O:\exists B_2\in\mathcal B_P:x+B_2\subseteq O\tag5,$$ By $(3)$, the condition $(4)$ is weaker than $(5)$ and hence there are more $\tau(\overline p)$-open sets than $\tau(P)$-open sets, i.e. $\tau(P)$ is coarser than $\tau(\overline p)$. This should be correct now, right? – 0xbadf00d Jan 04 '21 at 18:28
  • I don't agree with your $B=1\cdot U_{\overline p}= C(X,Y)$, but the next line shows that every element of $\mathcal B_{\overline p}$ is contained in an element of $\mathcal B_P$. For the rest of the derivation, what is necessary is that every element of $\mathcal B_{P}$ contains an element of $\mathcal B_{\overline p}$ - otherwise you cannot go from $(5)$ to $(4)$. My comment above was not correct about this condition (the condition of when $\mathcal B_1$ is finer than $\mathcal B_2$). – s.harp Jan 06 '21 at 17:16

0 Answers0