Let $X$ be a non-empty set and let $\mathbb{C}^X$ be the space of functions $X\longrightarrow \mathbb{C}$. It's easy to define operations wich makes $\mathbb{C}^X$ into vector space. Let $H\subset X$ be a finite set ( $H\in\wp_{<\omega}(X)$). For some $\varepsilon>0$, define $$V_{H,\varepsilon} := \{ f\in \mathbb{C}^X \mid x\in H \Rightarrow \|f(x)\| \leqslant \varepsilon\}.$$
I wanna see if the set $$\mathscr B := \{ V_{H,\varepsilon} \in \wp(\mathbb C^X) \mid \varepsilon> 0, H\in \wp_{_{<\omega}}(X)\}$$ is a filter in $\mathbb C^X$. I guess that some hypothesis are missing, like $X$ beeing compact or whatever, having trouble to prove that $U\supset V\in \mathscr B\Rightarrow U\in \mathscr B$.
The collection $\mathcal{B}$ is only a filter base: if $V_{H,\varepsilon}$ and $V_{H',\varepsilon'}$ are two members, their intersection contains $V_{H \cup H', \min(\varepsilon, \varepsilon')} \in \mathcal{B}'$ and all members of $\mathcal{B}$ are non-empty (they all contain the constant $0$-function, e.g.).
These are in fact a standard filter base for the neighbourhoods for that $0$-function when $\Bbb C^X$ gets the product topology as a power of $\Bbb C$ and defines a TVS-topology in the usual way.