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I am not sure I understand the difference between nets and filterbases very clearly, esp since there convergence properties seem to coincide. Isn't a filterbase a kind of a net?

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A net is a function whose domain is a directed set; a filter base is a particular kind of directed set. These are closely related notions, but they’re clearly not quite the same animal.

Suppose that $\mathscr{B}$ is a filter base in a space $X$. If for each $B\in\mathscr{B}$ I pick a point $x_B\in X$, then the function $\nu:\mathscr{B}\to X:B\mapsto x_B$ is a net in $X$. Thus, from a filter base I can easily construct a net, and one that turns out to have similar convergence properties, but the net $\nu$ is not the filter base $\mathscr{B}$.

Conversely, if $\langle D,\le\rangle$ is a directed set, and $\nu:D\to X:d\mapsto\nu_d$ is a net in $X$, for each $d\in D$ let $B_d=\{\nu_e:d\le e\}$; then $\mathscr{B}=\{B_d:d\in D\}$ is a filter base in $X$. Again you’ll find that $\nu$ and $\mathscr{B}$ have similar convergence properties, but they are not the same thing.

It’s a bit like the relationship between equivalence relations and partitions: they’re not the same thing, but they have a very close relationship that allows each be constructed from the other. The relationship between nets and filter bases isn’t quite as close, however, because the two constructions that I gave above aren’t actually inverses of each other.

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  • $\begingroup$ "Suppose $\mathcal{B}$ is a filterbase in $X$. If for each $B \in \mathcal{B}$, we have $x_B \in X$, then $\mathcal{x}: \mathcal{B} \rightarrow X$ such that $f(B)= x_B$ is a net in $X$." In order to prove the above statement would it be enough to prove the following: (i)$\mathcal{B}$ is a directed set. (ii) For every $\mathcal{B}$ in $X$ converging to $b$, $\mathcal{x}$ converges to $b$.(iii) For every $\mathcal{x}$ in $X$ converging to $x$, $\mathcal{B}$ converges to $x$.? $\endgroup$ – Fatsho Feb 20 '13 at 12:14
  • $\begingroup$ For the quoted statement (which you misquote: the function was called $\nu$, not $x$) we only have to prove (i), as by definition a net is just a function with domain a directed set. And $\mathcal{B}$ is directed by reverse inclusion (the directedness is then just a restatement of the fact that $\mathcal{B}$ is a filterbase. The convergence statements make no sense to me. $\endgroup$ – Henno Brandsma Feb 20 '13 at 18:29
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An addition to Brian's answer:

One nice property of filterbases: they're internal: families of subsets of the space itself. A net used any external index set we like, as long as it is directed, of course. So there is a set of all filters / filterbases on a set $X$, while there is no set (it's a proper class) of all nets in $X$. Another nice property of filter(base)s: the notion of a finer filter(base) is natural, and corresponds to subsequences, essentially, while the notion of subnet is a bit more unnatural IMHO. You need some examples to see why it is chosen the way it is, and there are actually two different notions of subnet with some different properties.

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