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Let $\mathcal{B}$ a filter basis on a topological space $X$ and define $\Lambda=\{(a,A):a\in A\in\mathcal{B}\}$. Define the net $x:\Lambda\rightarrow X$ by $x(a,A)=a$.

I need to prove that the filter generated by $\mathcal{B}$ is an ultrafilter $\Leftrightarrow (x_{\lambda})_{\lambda\in{\Lambda}}$ is an universal net.

What I know:

  • A filter $\mathcal{F}$ in $X$ (assuming $\leq$ as the partial order of $\Lambda$) is a subset of $\Lambda$ not empty satisfying:

$\qquad(1)\,\forall\, x,y\in\mathcal{F},\,\exists\, z\in\mathcal{F}$ such that $z\leq x$ and $z\leq y$

$\qquad(2)\,\forall\,x\in\mathcal{F}$ and $\forall\,y\in\Lambda,\quad x\leq y\implies y\in\mathcal{F}$

  • An ultrafilter is a maximal filter, that is, does not exists filter $\mathcal{F'}$ in $X$ with $\mathcal{F}\subset\mathcal{F'}$ (strict inclusion)

  • A universal net is a net such that, for any $A\subset X$, the net is eventually in $A$ or eventually in $X\setminus A$

I write these definitions and simply nothing comes to me. Can someone help?

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  • $\begingroup$ A filter on $X$ in this context is a non-empty collection of subsets of $X$ such that $\emptyset \notin \mathcal{F}$, $\mathcal{F}$ is closed under finite intersections ($A, B \in \mathcal{F} \implies A \cap B \in \mathcal{F}$ ) and under supersets (i.e. $A \in \mathcal{F} \land A \subseteq B \implies B \in \mathcal{F}$). $\endgroup$ – Henno Brandsma Jun 13 '19 at 15:55
  • $\begingroup$ I define the direction (not partial order) $\le$ on $\Lambda$ in my answer. $\endgroup$ – Henno Brandsma Jun 13 '19 at 15:57
  • $\begingroup$ A non-empty collection of subsets $\mathcal{B}$ of $X$ is called a filter base when $\emptyset \notin \mathcal{B}$, $\forall B_1,B_2 \in \mathcal{B}: \exists B_3 \in \mathcal{B}: B_3 \subseteq B_1 \cap B_2$. The filter $\mathcal{F}$ generated by $\mathcal{B}$ is $\{A \subseteq X: \exists B \in \mathcal{B}: B \subseteq A \}$ and one easily checks this is a filter in the sense of my other comment. $\endgroup$ – Henno Brandsma Jun 13 '19 at 16:05
  • $\begingroup$ You're confusing the notion of a filter (base) on $X$ with a more general one, of which the definition I have is a specialised version), namely that of a filter base in a poset (or sometimes even a lattice). For topological convergence we use the filters of subsets as I defined in the other comment. $\endgroup$ – Henno Brandsma Jun 13 '19 at 21:27
  • $\begingroup$ You actually don't define a partial order on $\Lambda$..... $\endgroup$ – Henno Brandsma Jun 15 '19 at 23:03
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To complete the definition of $\Lambda$, we have to define the direction relation $\le$:

$$\forall B_1,B_2 \mathcal{B}, \forall b_1 \in B, \forall b_2 \in B:\, (b_1,B_1) \le (b_2, B_2) \text{ iff } B_2 \subseteq B_1\tag{1}$$

and it's easy to check that this is a well-defined direction relation (not a partial order as two elements can be $\le$ each other and not equal; no antisymmetry) by the definition of a filter base: if $(b_1,B_1), (b_2,B_2) \in \Lambda$ we know that $B_1,B_2 \in \mathcal{B}$ so that there exists a $B_3 \in \mathcal{B}$ such that $B_3 \subseteq B_1 \cap B_2$ (definition of filter base) and pick $b_3 \in B_3$ (as all members of $\mathcal{B}$ are non-empty) and note that $(b_1,B_1) \le (b_3,B_3)$ and also $(b_2,B_2) \le (b_3,B_3)$ by definition (1). The fact that $\le$ is reflexive and transitive is very easy to check.

Lemma

If $\mathcal{B}$ is a filterbase and $\mathcal{F}$ is its generated filter and $\Lambda$, $\le$ and $x$ are as defined in the OP and as above, then $$\forall A \subseteq X: x \text{ is eventually in } A \iff A \in \mathcal{F}$$

proof: $\Rightarrow$: As $x$ is eventually in $A$ there is some $(b_0,B_0) \in \Lambda$ such that $$\forall (b,B) \in \Lambda: (b,B) \ge (b_0,B_0) \implies x(b,B) \in A$$

Now, if $p \in B_0$ is arbitary, by definition $(1)$ (where the point part of the pair is irrelevant to the $\le$ relation) we have that $(p,B_0) \ge (b_0,B_0)$ so $p=x(p,B_0) \in A$. As $p$ was arbitary, we have shown $B_0 \subseteq A$ and so $A$ is in the filter $\mathcal{F}$ generated by $\mathcal{B}$, as $B_0 \in \mathcal{B}$ by definition.

$\Leftarrow$: $A \in \mathcal{F}$ so there is some $B_0 \in \mathcal{B}$ with $B_0 \subseteq A$. Now pick any $b_0 \in B_0$ so that $(b_0,B_0) \in \Lambda$ and if $(b,B) \ge (b_0,B_0)$ we know that $B \subseteq B_0$ by $(1)$ and also that $x(b,B) = b \in B \subseteq B_0 \subseteq A$, so $(b_0,B_0)$ shows that $x$ is eventually in $A$. QED.

Now the proof is finished by knowing that $\mathcal{F}$ is an ultrafilter on $X$ iff $$\forall A \subseteq X: A \in \mathcal{F} \lor X\setminus A \in \mathcal{F}$$

as we see here, e.g.

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