Neighbourhood system for strong operator topology

Question

In the original paper by Hausdorff, a topological space was specified by a neighbourhood system $\mathcal{N}_{x}$ that satisfies the following axioms:

1. $\mathcal{N}_{x}\neq \varnothing$;
2. $V \in \mathcal{N}_{x}$ implies $x \in V$;
3. $V \in \mathcal{N}_{x}$ and $W \supseteq V$ implies $W \in \mathcal{N}_{x}$;
4. $V \in \mathcal{N}_{x}$ and $W \in \mathcal{N}_{x}$ implies $V \cap W \in \mathcal{N}_{x};$
5. For every $V \in\mathcal{N}_{x}$ there is $U \in \mathcal{N}_{x}$ with $V \in \mathcal{N}_{y}$ for all $y \in U$.

In case these axioms are satisfied, there exists a unique topology $\mathcal{O}$, which is generated by the neighbourhood system $(\mathcal{N}_{x})_{x \in X}$: \begin{equation} \mathcal{O}:=\{U \subseteq X| U \in \mathcal{N}_{x} \; \; \text{for all} \; \; x \in U \}. \end{equation} So my question translates to: what choice of $\mathcal{N}_{x}$ makes $\mathcal{O}$ to be the strong operator topology? My initial guess was the following:

Let $U(\mathcal{H})$ denote the space of unitary operators on a separable Hilbert space $\mathcal{H}$ and let $U_0 \in U(\mathcal{H})$. We define: \begin{equation} V_{f}(U_0,r):=\{U \in U(\mathcal{H})| \; \; ||U_0(f)-U(f)||<r \}, \; \; f \in \mathcal{H}, r>0. \end{equation} Then, I tried to construct the neighbourhood system $\mathcal{N}_{U_0}$ as follows: \begin{equation} \mathcal{N}_{U_0}:=\{V_f(U_0,r)|f \in \mathcal{H},r >0 \}. \end{equation} However, this does not seem to work, as I can't prove the axioms 3,4,5. So, perhaps, my choice of $\mathcal{N}_{U_0}$ is wrong. What would be the right choice, or how to prove 3,4,5? I will show my proof for $1,2$.

For $1$, we have: $\mathcal{N}_{U_0} \neq \varnothing$, as $U_0 \in \mathcal{N}_{U_0}$ always. This follows from the very definition of $V_f(U_0,r)$, it contains $U_0$ for any choice of $f$ and $r$.

For $2$, we have: Suppose $V \in \mathcal{N}_{U_0}$. Then, there exists $f \in \mathcal{H}$ and $r>0$, such that $V=V_{f}(U_0,r)$. But as we know that $U_0 \in V_f(U_0,r)$, we can conclude that $U_0 \in V$.

For 3,4,5, I do not know how to proceed.

Answer 1

$\newcommand{\L}{\mathcal{L}}\newcommand{\N}{\mathcal{N}}$I use the following definition of strong operator topology:

For two normed linear spaces $X,Y$, we topologise the set $\L(X,Y)$ of bounded linear operators $X\to Y$ by the subbase containing all set of the following form, for $x\in X$, $\epsilon>0$ and $T\in\L(X,Y)$: $$\N_{x,\epsilon}(T):=\{S\in\L(X,Y):\|(S-T)x\|<\epsilon\}$$

Now, given any topology, we can very simply extract a defining neighbourhood system by just declaring: $$\tag{$\ast$}\N_T:=\{U\subseteq\L(X,Y):\exists V\subseteq U,\,T\in V\text{ and $V$ is open}\}$$

In fact, I believe that is the only way to define $\N_T$ such that it is a neighbourhood system and such that it generates the same topology.

Notice that it is not true that any choice of local base $\N_T$ at $T\in\L(X,Y)$ will assemble to a neighbourhood system $(\N_T)_{T\in\L(X,Y)}$. Axiom $(3)$ poses a big problem: our neighbourhoood system at $T$ should have to contain the subbasic sets $\N_{x,\epsilon}(T)$ but then axiom $(3)$ implies $\N_T$ contains all containers of $\N_{x,\epsilon}(T)$... (for some $x$ and some $\epsilon$) meaning $\N_T$ must contain all neighbourhoods of $T$. So you can't really do better than $(\ast)$.

You can write:

$$\mathcal{N}_T:=\{U\subseteq\mathcal{L}(X,Y):\text{ For some $n\in\Bbb N$, $x_1,x_2,\cdots,x_n\in X$, $\epsilon>0$ we have }\bigcap_{j=1}^n\mathcal{N}_{x_j,\epsilon}(T)\subseteq U\}$$

But this is essentially exactly the same as the subbase definition! You don't gain anything new.

Your own choice should be wrong in general since there will exist neighbourhoods not of the form $V_f(U_0;r)$ and such neighbourhoods must be contained in $\N_{U_0}$. Indeed, the whole space $\L(X,Y)$ is in $\N_{U_0}$ but is essentially never of the form $V_f(U_0;r)$, if $\N_{U_0}$ is part of a genuine neighbourhood system.

By the way I think you made a typo when defining $\mathcal{O}$. It should be: $$\mathcal{O}:=\{U\subseteq X:U\in\N_x\,\forall x\in U\}$$