# About $\sigma$ strong$^*$-functionals and seminorms

I'm reading the book "Theory of operator algebras" by Takesaki. In this book, the $$\sigma$$-strong$$^*$$ topology on the space $$B(H)$$ (bounded operators on the Hilbert space $$H$$) is defined (see def 2.3 p68) to be the locally convex topology generated by the seminorms

$$B(H)\ni x \mapsto \left(\sum_{n=1}^\infty \|x\xi_n\|^2 + \|x^*\xi_n\|^2\right)^{1/2}.$$I am trying to understand the proof of lemma 2.4: Question:

Why does it suffice to use only one seminorm? Shouldn't we have something like $$\sum_{k=1}^m \left(\sum_{n=1}^\infty (\|x(\xi_k)_n\|^2+ \|x^*(\xi_k)_n\|^2\right)^{1/2} \le 1 \implies |\omega(x)| \le 1$$ where $$\xi_1, \dots, \xi_m$$ are sequences in $$H$$.

So, we are dealing with locally convex topological vector spaces. I think, in general, given a family of seminorms, you would need to consider the finite intersections of the open balls they form, see wikipedia article. So a basic open set about $$0$$ is of the form $$\{ x : p(x) where $$F$$ is a finite subset of the set $$\mathcal P$$ of generating seminorms, and $$r>0$$. However, in the special case of the $$\sigma$$-strong$$^*$$-toplogy, this isn't necessary.
In our case, the seminorms are of the form $$p_{\xi}(x) = \Big( \sum_{n=1}^\infty \|x\xi_n\|^2 + \|x^*\xi_n\|^2 \Big)^{1/2},$$ for any sequence $$\xi=(\xi_n)$$ with $$\sum_n \|\xi_n\|^2<\infty$$. We can always rescale, so it suffices to take $$r=1$$ above. Now notice the special property: given $$\xi,\eta$$, with $$\gamma$$ the union of the sequences $$\xi,\eta$$ (say $$\gamma_{2n} = \xi_n, \gamma_{2n-1} = \eta_n$$) then if $$p_\gamma(x) < 1$$ then certainly $$p_\xi(x)<1$$ and $$p_\eta(x)<1$$. The same argument applies to finite families.
We conclude that for our $$\mathcal P$$, for any basic open set $$U$$ about $$0$$ there is some $$p\in\mathcal P$$ so that $$\{ x : p(x)<1 \}$$ is contained in $$U$$.
A linear functional $$f$$ is continuous if and only if $$f$$ is bounded on some open set about $$0$$. So, with $$\mathcal P$$ having our property, f is continuous if and only if there is some $$p\in\mathcal P$$ with $$\{ f(x) : p(x)<1 \}$$ bounded. Again by rescaling, equivalently, there is some $$p\in\mathcal P$$ with $$|f(x)|\leq 1$$ for all $$x$$ with $$p(x)<1$$.
Finally, a further standard rescaling argument shows that you can use $$p(x)\leq 1$$ instead of $$p(x)<1$$.