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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?

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  • $\begingroup$ What is your book's definition of a "net based on a filter"? I would imagine that definition might answer your question. $\endgroup$ – Eric Wofsey Jun 5 at 3:09
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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 $\langle A,\supseteq\rangle$.

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  • $\begingroup$ thanks sir for your kindness $\endgroup$ – ILoveMath Jun 6 at 0:08
  • $\begingroup$ @ILoveMath: You’re very welcome. $\endgroup$ – Brian M. Scott Jun 6 at 5:09
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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 $(x_i) \to y$.

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