Any filter is contained in a ultrafilter - Proof Explanation

Question

Theorem: Let F be a filter $\implies \exists \;U$ ultrafilter s.t. $F\subset U$

Notation:$Flt^+(M) = \{F: F $ is a semi-filter in $M\}$ where a semi-filter is a filter that may contain empty sets. (sorry, I'm unsure if "semi-filter" is the correct term, this is a direct translation of the portuguese term "quase-filtro")

Let $\mathscr{F}$ be a filter on M and $S \subset M$ s.t. $S \not \in \mathscr{F}. $

Let $\mathscr{A}=\{X \in Flt^+(M): S\not \in X \supset \mathscr{F} \}$

The set $\mathscr{A}$ is ordered in inductive, so by Zorn's lemma there's a maximal element $\mathscr{G}$ in $\mathscr{A}$.

Obviously $\mathscr{G}$ countains $\mathscr{F}$. We now must show that $\mathscr{G}$ is an ultrafilter.

We know that $U$ is an ultrafilter $\iff A\cup B\in U \implies A\in U$ or $B\in U$, so by contradiction let's suppose $A\cup B \in \mathscr{G}$ and $A \not \in \mathscr{G}, B \not \in \mathscr{G}$

Let $\mathscr{G}_A$ be the semi-filter generated by $\mathscr{G}\cup \{A\}$. By the maximality of $\mathscr{G}$, we have that $S \in \mathscr{G}_A$ (otherwise $\mathscr{G}_A$ would be a filter contained in $\mathscr{A}$ "bigger" than $\mathscr{G}$). Defining $\mathscr{G}_B$ likewise we get that $S \in \mathscr{G}_B$ as well.

Now what I can't understand is the following:

$S \in \mathscr{G}_A$ and $S \in \mathscr{G}_B \implies \exists C_1, C_2 \in \mathscr{G}$ s.t. $S \supset C_1\cap A, \quad S \supset C_2 \cap B.$

$\implies S \supset C_1\cap C_2\cap A, \quad S \supset C_1\cap C_2\cap B$

(why do the sets $C_1,C_2$ exist???)

therefore $S \supset (C_1\cap C_2\cap A)\cup (C_1 \cap C_2\cap B) = (A\cup B)\cap (C_1 \cap C_2) \in \mathscr{G}$

$\implies S \in \mathscr{G}$ which is a contradiction.

$\blacksquare$

I understand everything but the marked part, I suppose there would be a contradiction with the maximality of $\mathscr{G}$ if there wasn't such sets $C_1$ and $C_2$ but I couldn't prove it. Can anyone explain me that?

Answer 1

Ok, I believe I understood it. (Please, correct me if I'm wrong)

Definition: (1) $A, B \in \mathbb{F} \implies A\cap B \in \mathbb{F}$

(2) $B \supset A \in \mathbb{F} \implies B \in \mathbb{F}$

(3) $\emptyset \not \in \mathbb{F}$

if Properties (1), (2) and (3) are satisfied, $\mathbb{F}$ is a filter. If (1) and (2) are satisfied, $\mathbb{F}$ is a semi-filter.

Suppose there are no set $C\in \mathscr{G}$ such that $A\cap C \subset S$. That is, for every $X \in \mathscr{G},\exists y \in (X\cap A)|S. (X\cap A \not =\emptyset$ since $\emptyset \subset S)$

Claim:if that's true, then $\mathscr{G}_A|S$ is a semi-filter.

(1)Let $B, C \in \mathscr{G}_A|S\implies B \cap C \in \mathscr{G}_A$ and $B\cap C \not \subset S\implies B\cap C \not = S\implies B\cap C \in \mathscr{G}_A|S$

(2)Since $\forall X\in \mathscr{G}, X\cap A\not \subset S \implies X \not \subset S$ and $A \not \subset S$ therefore S does not contain any sets in $\mathscr{G}_A$

so $B\supset A \in \mathscr{G}_A|S \implies B\in \mathscr{G}_A$ and $B \not = S\implies B\in \mathscr{G}_A|S$.

Therefore $\mathscr{G}_A|S$ is a semi filter smaller than $\mathscr{G}_A$ that contains $\mathscr{G}\cap A$ which is a contradiction since $\mathscr{G}_A$ is supposed to be the smallest semi-filter containing such sets. $\square$

Answer 2

The following lemma is implicit in your proof and shows the existence of the $C_1,C_2$:

If $\mathcal{F}$ is a quasi-filter on $X$ and $A \subseteq X$ the smallest quasi-filter that contains $\mathcal{F}$ and $A$ as well equals : $$\mathcal{F}_A = \{B \subseteq X\mid \exists B' \in \mathcal{F}: B'\cap A \subseteq B\}$$

Let's check all this: Firstly, $\mathcal{F}_A$ is a quasi-filter:

1. If $B$ is in the set described on the right hand side and $B_1 \supseteq B$, then the same $B'$ that witnesses that $B$ is in the set, also does for $B_1$. So the superset axiom is fulfilled.

2. If $B_1, B_2 \in \mathcal{F}_A$ find $B'_1 \in \mathcal{F}$ such that $B'_1 \cap A \subseteq B_1$ and likewise a $B'_2 \in \mathcal{F}$ such that $B'_2 \cap A \subseteq B_2$. Then $B'_1 \cap B'_2 \in \mathcal{F}$ and $$(B'_1 \cap B'_2) \cap A = (B'_1 \cap A) \cap (B'_2 \cap A) \subseteq B_1 \cap B_2$$ so that $B_1 \cap B_2 \in \mathcal{F}_A$ as well, as witnessed by $B'_1 \cap B'_2$.

3. If $B \in \mathcal{F}$ then surely $B \cap A \subseteq B$ so it is itself a witness that $B \in \mathcal{F}_A$. So $\mathcal{F} \subseteq \mathcal{F}_A$.

4. Take $X \in \mathcal{F}$ then $A \cap X \subseteq A$, so $X$ (which is in any non-empty quasi-filter on $X$) witnesses that $A \in \mathcal{F}_A$ too.

5. If $\mathcal{G}$ is any quasi-filter such that $\mathcal{F} \subseteq \mathcal{G}$ and $A \in \mathcal{G}$, then for any $B \in \mathcal{F}_A$: pick $B' \in \mathcal{F}$ as promised in the definition, so that $B' \cap A \subseteq B$. But then both $B'$ and $A$ are in $\mathcal{G}$ as well, and hence their intersection, and thus also $B$ it superset, using both axioms for a quasi-filter. So $\mathcal{F}_A \subseteq \mathcal{G}$, and as $\mathcal{G}$ was arbitary, we have shown minimality.

6. As a last note, if $\mathcal{F}$ is a (proper) filter, $\mathcal{F}_A$ will also be iff $A$ intersects all members of $\mathcal{F}$.