# The filter generated by the filter basis $\mathcal{B}$ is an ultrafilter iff $(x_{\lambda})_{\lambda\in\Lambda}$ is an universal net

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?

• 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. $A \in \mathcal{F} \land A \subseteq B \implies B \in \mathcal{F}$). – Henno Brandsma Jun 13 '19 at 15:55
• I define the direction (not partial order) $\le$ on $\Lambda$ in my answer. – Henno Brandsma Jun 13 '19 at 15:57
• 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. – Henno Brandsma Jun 13 '19 at 16:05
• 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. – Henno Brandsma Jun 13 '19 at 21:27
• You actually don't define a partial order on $\Lambda$..... – Henno Brandsma Jun 15 '19 at 23:03

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.