# Net based on a filter

Assume a topological space X, t. I read a proposition in my textbook:
A filter A $$\to$$ y if and only if every net {$$s_a$$}, a $$\in$$ A, based on A also converges to y.

What are the net elements $$s_A$$ in this case? Are they elements of the topological space X or are they elements of the filter A? Also, why is the index set, a $$\in$$ A, when A is a filter and not a directed set?

• What is your book's definition of a "net based on a filter"? I would imagine that definition might answer your question. – Eric Wofsey Jun 5 at 3:09

Every filter is a directed set. Here, for instance, $$\langle A,\supseteq\rangle$$ is a directed set: you can easily check that the conditions that make $$A$$ a filter imply those that make $$\langle A,\supseteq\rangle$$ a directed set. The $$s_a$$ are points of $$X$$: for each $$a\in A$$ we choose any point $$s_a\in a$$. Then $$\langle s_a:a\in A\rangle$$ is a net indexed by the directed set $$\langle A,\supseteq\rangle$$.
The canonical net based on the filter $$A$$ on $$X$$ is based on the directed set
$$I:=\{(a,x) \in A \times X\mid x \in a, a \in A\}$$ ordered by $$(a,x) \le (a',x') \iff a' \subseteq a$$
which is easily seen to be a preorder that makes $$I$$ a directed set. The net itself is defined by the map $$f: (a,x) \in I \to x$$, the second projection.
Now if $$A \to y$$ in $$X$$, this net $$I \to X$$ also converges to $$y$$: let $$O$$ be an open neighbourhood of $$y$$ in $$X$$. Then as $$A \to y$$, we known $$O \in A$$. Now let $$x=p$$ be any point in $$A$$ (a filter has non-empty members so that's no problem), and then $$i_0 := (O,p)$$ is in $$I$$ by definition and if $$(a,x) \ge i_0=(O,p)$$ we know that $$x \in a (\in A)$$ and $$a \subseteq O$$ so $$f(a,x)=x \in O$$ and so by definition of net-convergence $$(x_i) \to y$$.