Does a basis for the uniform operator topology look like this? The set of all balls around all points T in B(X,Y) of all radii epsilon: $$\mathcal{B} = \bigcup_{T \in B(X,Y)} B_T(\varepsilon)$$ where $$B_T(\varepsilon) = \{ T^\prime \mid \sup_{\| f \| \leq 1} \| (T - T^\prime)f \| < \varepsilon} $$ for all $\varepsilon$ in $\mathbb{R}$?
the where case: $$B_T( \varepsilon) = \{ \tilde{T} \mid \sup_{\| f \| \leq 1} \vert (T - T^\prime) f \vert < \varepsilon \} $$
That mod inside the sup expression should've been $\| \cdot \|_Y$
What you said in words is correct, but it's not what you wrote. $$\mathcal{B} = \{B_{\varepsilon}(T)\,:\,T \in B(X,Y),\,\varepsilon \gt 0\}$$ would be what I'd write.
I wanted to spell out the bit that tells me what the metric is. Because next I'm going to have to worry about a basis for the strong operator topology : )
Thanks, btw.
That was a typo. This $\bigcup \{ B_T (\varepsilon) \}$ is what I actually meant to write.
Before worrying about the strong operator topology explicitly: given a vector space $X$ and a family $\mathscr{P}$ of semi-norms on $X$ can you write down a basis for the topology?
By the way: are the balls really written like that in your class? It strikes me as odd to emphasize the radius more than the center. What I usually see is either $B_{r}(x)$ or $B(x,r)$.
The strong operator topology, for instance, is the topology induced by the family $$\mathscr{P} = \{p_{x}\,:\,x \in X\}$$ where $p_{x}(T) = \|Tx\|_{Y}$.
I think there is a typo in the lecture notes. They define the strong operator topology on $B(X,Y)$ to be the weakest topology for which the evaluation map $L \mapsto Lx$ is continuous for every $x$. It should be $L \mapsto \| Lx \|_Y$ for the norm on $Y$. But then there is only one norm and no family of semi-norms.
