The question struck me while going through one paper which mentions Zabreiko lemma which says 'Every countably subadditive seminorm on a Banach space is continuous.' Where seminorm $p$ is said to be countably additive when $p(\displaystyle \sum_n z_n) \leq \displaystyle \sum_n p(z_n)$.
This means there do exists some seminorms which are not countably subadditive. I am struck at this point. Am I thinking something mad here. I searched on internet about countable subadditivity property of norm or seminorm. But couldn't find any relevant results. I feel triangle inequality should imply this property for both but then the statement in lemma should have been for every seminorm which is not. Please help.
Every continuous seminorm is countably subadditive. For we have $$p\biggl(\sum_n z_n\biggr) = \lim_{k \to \infty} p\biggl(\sum_{n \leqslant k} z_n\biggr) \leqslant \lim_{k \to \infty} \sum_{n \leqslant k} p(z_n) = \sum_n p(z_n)$$ by continuity, the triangle inequality, and the definition of the sum of an infinite series.
But discontinuous seminorms exist, and those need not be countably subadditive (unless one rejects enough of the axiom of choice, there are for example discontinuous seminorms on every infinite-dimensional Banach space). Consider on $\ell^2(\mathbb{N})$ a discontinuous linear functional $f$ such that $f(e_n) = 0$ for all $n \in \mathbb{N}$ but $$f\biggl(\sum_n \frac{1}{n} e_n\biggr) \neq 0\,.$$ Then $\lvert f\rvert$ is a seminorm, but by construction it is not countably subadditive.